Calculus Examples

$f (x) = e^{- x^{4}}$

Find the first derivative.

Tap for more steps...

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u$ as $- x^{4}$ .

$f' (x) = \frac{d}{d u} (e^{u}) \frac{d}{d x} (- x^{4})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$f' (x) = e^{u} \frac{d}{d x} (- x^{4})$

Replace all occurrences of $u$ with $- x^{4}$ .

$f' (x) = e^{- x^{4}} \frac{d}{d x} (- x^{4})$

Differentiate.

Tap for more steps...

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f' (x) = e^{- x^{4}} (- \frac{d}{d x} x^{4})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = e^{- x^{4}} (- (4 x^{3}))$

Multiply $4$ by $- 1$ .

$f' (x) = e^{- x^{4}} (- 4 x^{3})$

Simplify.

Tap for more steps...

Reorder the factors of $e^{- x^{4}} (- 4 x^{3})$ .

$f' (x) = - 4 e^{- x^{4}} x^{3}$

Reorder factors in $- 4 e^{- x^{4}} x^{3}$ .

$f' (x) = - 4 x^{3} e^{- x^{4}}$

Find the second derivative.

Tap for more steps...

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{3} e^{- x^{4}}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{3} e^{- x^{4}}]$ .

$f'' (x) = - 4 \frac{d}{d x} (x^{3} e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{3}$ and $g (x) = e^{- x^{4}}$ .

$f'' (x) = - 4 (x^{3} \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u$ as $- x^{4}$ .

$f'' (x) = - 4 (x^{3} (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - 4 (x^{3} (e^{u} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Replace all occurrences of $u$ with $- x^{4}$ .

$f'' (x) = - 4 (x^{3} (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Differentiate.

Tap for more steps...

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f'' (x) = - 4 (x^{3} (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = - 4 (x^{3} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Multiply $4$ by $- 1$ .

$f'' (x) = - 4 (x^{3} (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Multiply $x^{3}$ by $x^{3}$ by adding the exponents.

Tap for more steps...

Move $x^{3}$ .

$f'' (x) = - 4 (x^{3} x^{3} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - 4 (x^{3 + 3} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Add $3$ and $3$ .

$f'' (x) = - 4 (x^{6} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f'' (x) = - 4 (x^{6} (- 4 \cdot e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{3}))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = - 4 (x^{6} (- 4 e^{- x^{4}}) + e^{- x^{4}} (3 x^{2}))$

Simplify.

Tap for more steps...

Apply the distributive property.

$f'' (x) = - 4 (x^{6} (- 4 e^{- x^{4}})) - 4 (e^{- x^{4}} (3 x^{2}))$

Combine terms.

Tap for more steps...

Multiply $- 4$ by $- 4$ .

$f'' (x) = 16 (x^{6} (e^{- x^{4}})) - 4 (e^{- x^{4}} (3 x^{2}))$

Multiply $3$ by $- 4$ .

$f'' (x) = 16 x^{6} e^{- x^{4}} - 12 e^{- x^{4}} x^{2}$

Reorder terms.

$f'' (x) = 16 e^{- x^{4}} x^{6} - 12 e^{- x^{4}} x^{2}$

Reorder factors in $16 e^{- x^{4}} x^{6} - 12 e^{- x^{4}} x^{2}$ .

$f'' (x) = 16 x^{6} e^{- x^{4}} - 12 x^{2} e^{- x^{4}}$

Find the third derivative.

Tap for more steps...

By the Sum Rule, the derivative of $16 x^{6} e^{- x^{4}} - 12 x^{2} e^{- x^{4}}$ with respect to $x$ is $\frac{d}{d x} [16 x^{6} e^{- x^{4}}] + \frac{d}{d x} [- 12 x^{2} e^{- x^{4}}]$ .

$f''' (x) = \frac{d}{d x} (16 x^{6} e^{- x^{4}}) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Evaluate $\frac{d}{d x} [16 x^{6} e^{- x^{4}}]$ .

Tap for more steps...

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{6} e^{- x^{4}}$ with respect to $x$ is $16 \frac{d}{d x} [x^{6} e^{- x^{4}}]$ .

$f''' (x) = 16 \frac{d}{d x} (x^{6} e^{- x^{4}}) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{6}$ and $g (x) = e^{- x^{4}}$ .

$f''' (x) = 16 (x^{6} \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u_{1}$ as $- x^{4}$ .

$f''' (x) = 16 (x^{6} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f''' (x) = 16 (x^{6} (e^{u_{1}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Replace all occurrences of $u_{1}$ with $- x^{4}$ .

$f''' (x) = 16 (x^{6} (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f''' (x) = 16 (x^{6} (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f''' (x) = 16 (x^{6} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x^{6})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 6$ .

$f''' (x) = 16 (x^{6} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Multiply $4$ by $- 1$ .

$f''' (x) = 16 (x^{6} (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Multiply $x^{6}$ by $x^{3}$ by adding the exponents.

Tap for more steps...

Move $x^{3}$ .

$f''' (x) = 16 (x^{3} x^{6} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f''' (x) = 16 (x^{3 + 6} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Add $3$ and $6$ .

$f''' (x) = 16 (x^{9} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) + \frac{d}{d x} (- 12 x^{2} e^{- x^{4}})$

Evaluate $\frac{d}{d x} [- 12 x^{2} e^{- x^{4}}]$ .

Tap for more steps...

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x^{2} e^{- x^{4}}$ with respect to $x$ is $- 12 \frac{d}{d x} [x^{2} e^{- x^{4}}]$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 \frac{d}{d x} (x^{2} e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{- x^{4}}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u_{2}$ as $- x^{4}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{u_{2}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Replace all occurrences of $u_{2}$ with $- x^{4}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x^{2}))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} (2 x))$

Multiply $4$ by $- 1$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{2} (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} (2 x))$

Multiply $x^{2}$ by $x^{3}$ by adding the exponents.

Tap for more steps...

Move $x^{3}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{3} x^{2} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (2 x))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{3 + 2} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (2 x))$

Add $3$ and $2$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{5} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (2 x))$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}}) + e^{- x^{4}} (6 x^{5})) - 12 (x^{5} (- 4 e^{- x^{4}}) + e^{- x^{4}} (2 x))$

Simplify.

Tap for more steps...

Apply the distributive property.

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}})) + 16 (e^{- x^{4}} (6 x^{5})) - 12 (x^{5} (- 4 e^{- x^{4}}) + e^{- x^{4}} (2 x))$

Apply the distributive property.

$f''' (x) = 16 (x^{9} (- 4 e^{- x^{4}})) + 16 (e^{- x^{4}} (6 x^{5})) - 12 (x^{5} (- 4 e^{- x^{4}})) - 12 (e^{- x^{4}} (2 x))$

Combine terms.

Tap for more steps...

Multiply $- 4$ by $16$ .

$f''' (x) = - 64 (x^{9} (e^{- x^{4}})) + 16 (e^{- x^{4}} (6 x^{5})) - 12 (x^{5} (- 4 e^{- x^{4}})) - 12 (e^{- x^{4}} (2 x))$

Multiply $6$ by $16$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 96 (e^{- x^{4}} (x^{5})) - 12 (x^{5} (- 4 e^{- x^{4}})) - 12 (e^{- x^{4}} (2 x))$

Multiply $- 4$ by $- 12$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 96 e^{- x^{4}} x^{5} + 48 (x^{5} (e^{- x^{4}})) - 12 (e^{- x^{4}} (2 x))$

Multiply $2$ by $- 12$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 96 e^{- x^{4}} x^{5} + 48 x^{5} e^{- x^{4}} - 24 (e^{- x^{4}} (x))$

Add $96 e^{- x^{4}} x^{5}$ and $48 x^{5} e^{- x^{4}}$ .

Tap for more steps...

Move $e^{- x^{4}}$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 96 x^{5} e^{- x^{4}} + 48 x^{5} e^{- x^{4}} - 24 e^{- x^{4}} x$

Add $96 x^{5} e^{- x^{4}}$ and $48 x^{5} e^{- x^{4}}$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 144 x^{5} e^{- x^{4}} - 24 e^{- x^{4}} x$

Reorder terms.

$f''' (x) = - 64 e^{- x^{4}} x^{9} + 144 e^{- x^{4}} x^{5} - 24 e^{- x^{4}} x$

Reorder factors in $- 64 e^{- x^{4}} x^{9} + 144 e^{- x^{4}} x^{5} - 24 e^{- x^{4}} x$ .

$f''' (x) = - 64 x^{9} e^{- x^{4}} + 144 x^{5} e^{- x^{4}} - 24 x e^{- x^{4}}$

Find the fourth derivative.

Tap for more steps...

By the Sum Rule, the derivative of $- 64 x^{9} e^{- x^{4}} + 144 x^{5} e^{- x^{4}} - 24 x e^{- x^{4}}$ with respect to $x$ is $\frac{d}{d x} [- 64 x^{9} e^{- x^{4}}] + \frac{d}{d x} [144 x^{5} e^{- x^{4}}] + \frac{d}{d x} [- 24 x e^{- x^{4}}]$ .

$f^{4} (x) = \frac{d}{d x} (- 64 x^{9} e^{- x^{4}}) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Evaluate $\frac{d}{d x} [- 64 x^{9} e^{- x^{4}}]$ .

Tap for more steps...

Since $- 64$ is constant with respect to $x$ , the derivative of $- 64 x^{9} e^{- x^{4}}$ with respect to $x$ is $- 64 \frac{d}{d x} [x^{9} e^{- x^{4}}]$ .

$f^{4} (x) = - 64 \frac{d}{d x} (x^{9} e^{- x^{4}}) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{9}$ and $g (x) = e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{9} \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u_{1}$ as $- x^{4}$ .

$f^{4} (x) = - 64 (x^{9} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 64 (x^{9} (e^{u_{1}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Replace all occurrences of $u_{1}$ with $- x^{4}$ .

$f^{4} (x) = - 64 (x^{9} (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f^{4} (x) = - 64 (x^{9} (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f^{4} (x) = - 64 (x^{9} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x^{9})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 9$ .

$f^{4} (x) = - 64 (x^{9} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Multiply $4$ by $- 1$ .

$f^{4} (x) = - 64 (x^{9} (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Multiply $x^{9}$ by $x^{3}$ by adding the exponents.

Tap for more steps...

Move $x^{3}$ .

$f^{4} (x) = - 64 (x^{3} x^{9} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{4} (x) = - 64 (x^{3 + 9} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Add $3$ and $9$ .

$f^{4} (x) = - 64 (x^{12} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + \frac{d}{d x} (144 x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Evaluate $\frac{d}{d x} [144 x^{5} e^{- x^{4}}]$ .

Tap for more steps...

Since $144$ is constant with respect to $x$ , the derivative of $144 x^{5} e^{- x^{4}}$ with respect to $x$ is $144 \frac{d}{d x} [x^{5} e^{- x^{4}}]$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 \frac{d}{d x} (x^{5} e^{- x^{4}}) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{5}$ and $g (x) = e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u_{2}$ as $- x^{4}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{u_{2}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Replace all occurrences of $u_{2}$ with $- x^{4}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x^{5})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Multiply $4$ by $- 1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{5} (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Multiply $x^{5}$ by $x^{3}$ by adding the exponents.

Tap for more steps...

Move $x^{3}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{3} x^{5} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{3 + 5} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Add $3$ and $5$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) + \frac{d}{d x} (- 24 x e^{- x^{4}})$

Evaluate $\frac{d}{d x} [- 24 x e^{- x^{4}}]$ .

Tap for more steps...

Since $- 24$ is constant with respect to $x$ , the derivative of $- 24 x e^{- x^{4}}$ with respect to $x$ is $- 24 \frac{d}{d x} [x e^{- x^{4}}]$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 \frac{d}{d x} (x e^{- x^{4}})$

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x \frac{d}{d x} (e^{- x^{4}}) + e^{- x^{4}} \frac{d}{d x} (x))$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x^{4}$ .

Tap for more steps...

To apply the Chain Rule, set $u_{3}$ as $- x^{4}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (\frac{d}{d u (3)} (e^{u_{3}}) \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x))$

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{u_{3}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x))$

Replace all occurrences of $u_{3}$ with $- x^{4}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{- x^{4}} \frac{d}{d x} (- x^{4})) + e^{- x^{4}} \frac{d}{d x} (x))$

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{- x^{4}} (- \frac{d}{d x} x^{4})) + e^{- x^{4}} \frac{d}{d x} (x))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \frac{d}{d x} (x))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{- x^{4}} (- (4 x^{3}))) + e^{- x^{4}} \cdot 1)$

Multiply $4$ by $- 1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x (e^{- x^{4}} (- 4 x^{3})) + e^{- x^{4}} \cdot 1)$

Raise $x$ to the power of $1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{3} x (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \cdot 1)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{3 + 1} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \cdot 1)$

Add $3$ and $1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (e^{- x^{4}} \cdot (- 4)) + e^{- x^{4}} \cdot 1)$

Move $- 4$ to the left of $e^{- x^{4}}$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 \cdot e^{- x^{4}}) + e^{- x^{4}} \cdot 1)$

Multiply $e^{- x^{4}}$ by $1$ .

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}}) + e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}}) + e^{- x^{4}})$

Simplify.

Tap for more steps...

Apply the distributive property.

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}})) - 64 (e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}}) + e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}}) + e^{- x^{4}})$

Apply the distributive property.

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}})) - 64 (e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}})) + 144 (e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}}) + e^{- x^{4}})$

Apply the distributive property.

$f^{4} (x) = - 64 (x^{12} (- 4 e^{- x^{4}})) - 64 (e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}})) + 144 (e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Combine terms.

Tap for more steps...

Multiply $- 4$ by $- 64$ .

$f^{4} (x) = 256 (x^{12} (e^{- x^{4}})) - 64 (e^{- x^{4}} (9 x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}})) + 144 (e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Multiply $9$ by $- 64$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 576 (e^{- x^{4}} (x^{8})) + 144 (x^{8} (- 4 e^{- x^{4}})) + 144 (e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Multiply $- 4$ by $144$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 576 e^{- x^{4}} x^{8} - 576 (x^{8} (e^{- x^{4}})) + 144 (e^{- x^{4}} (5 x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Multiply $5$ by $144$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 576 e^{- x^{4}} x^{8} - 576 x^{8} e^{- x^{4}} + 720 (e^{- x^{4}} (x^{4})) - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Subtract $576 x^{8} e^{- x^{4}}$ from $- 576 e^{- x^{4}} x^{8}$ .

Tap for more steps...

Move $e^{- x^{4}}$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 576 x^{8} e^{- x^{4}} - 576 x^{8} e^{- x^{4}} + 720 e^{- x^{4}} x^{4} - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Subtract $576 x^{8} e^{- x^{4}}$ from $- 576 x^{8} e^{- x^{4}}$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 720 e^{- x^{4}} x^{4} - 24 (x^{4} (- 4 e^{- x^{4}})) - 24 e^{- x^{4}}$

Multiply $- 4$ by $- 24$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 720 e^{- x^{4}} x^{4} + 96 (x^{4} (e^{- x^{4}})) - 24 e^{- x^{4}}$

Add $720 e^{- x^{4}} x^{4}$ and $96 x^{4} e^{- x^{4}}$ .

Tap for more steps...

Move $e^{- x^{4}}$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 720 x^{4} e^{- x^{4}} + 96 x^{4} e^{- x^{4}} - 24 e^{- x^{4}}$

Add $720 x^{4} e^{- x^{4}}$ and $96 x^{4} e^{- x^{4}}$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 816 x^{4} e^{- x^{4}} - 24 e^{- x^{4}}$

Reorder terms.

$f^{4} (x) = 256 e^{- x^{4}} x^{12} - 1152 e^{- x^{4}} x^{8} + 816 e^{- x^{4}} x^{4} - 24 e^{- x^{4}}$

Reorder factors in $256 e^{- x^{4}} x^{12} - 1152 e^{- x^{4}} x^{8} + 816 e^{- x^{4}} x^{4} - 24 e^{- x^{4}}$ .

$f^{4} (x) = 256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 816 x^{4} e^{- x^{4}} - 24 e^{- x^{4}}$

The fourth derivative of $f (x)$ with respect to $x$ is $256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 816 x^{4} e^{- x^{4}} - 24 e^{- x^{4}}$ .

$256 x^{12} e^{- x^{4}} - 1152 x^{8} e^{- x^{4}} + 816 x^{4} e^{- x^{4}} - 24 e^{- x^{4}}$