Calculus Examples

$f (x) = e^{\sin (x)}$

Find the first derivative.

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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 $g (x) = \sin (x)$ .

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To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

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

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

Replace all occurrences of $u$ with $\sin (x)$ .

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

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f' (x) = e^{\sin (x)} \cos (x)$

Find the second derivative.

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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 $f (x) = e^{\sin (x)}$ and $g (x) = \cos (x)$ .

$f'' (x) = e^{\sin (x)} \frac{d}{d x} (\cos (x)) + \cos (x) \frac{d}{d x} (e^{\sin (x)})$

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

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 $g (x) = \sin (x)$ .

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To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

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

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

Replace all occurrences of $u$ with $\sin (x)$ .

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

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f'' (x) = e^{\sin (x)} (- \sin (x)) + \cos (x) e^{\sin (x)} \cos (x)$

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = e^{\sin (x)} (- \sin (x)) + \cos (x) \cos (x) e^{\sin (x)}$

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = e^{\sin (x)} (- \sin (x)) + \cos (x) \cos (x) e^{\sin (x)}$

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

$f'' (x) = e^{\sin (x)} (- \sin (x)) + {\cos (x)}^{1 + 1} e^{\sin (x)}$

Add $1$ and $1$ .

$f'' (x) = e^{\sin (x)} (- \sin (x)) + \cos^{2} (x) e^{\sin (x)}$

Reorder terms.

$f'' (x) = - e^{\sin (x)} \sin (x) + e^{\sin (x)} \cos^{2} (x)$

Find the third derivative.

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By the Sum Rule, the derivative of $- e^{\sin (x)} \sin (x) + e^{\sin (x)} \cos^{2} (x)$ with respect to $x$ is $\frac{d}{d x} [- e^{\sin (x)} \sin (x)] + \frac{d}{d x} [e^{\sin (x)} \cos^{2} (x)]$ .

$f''' (x) = \frac{d}{d x} (- e^{\sin (x)} \sin (x)) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

Evaluate $\frac{d}{d x} [- e^{\sin (x)} \sin (x)]$ .

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

$f''' (x) = - \frac{d}{d x} (e^{\sin (x)} \sin (x)) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

$f''' (x) = - (e^{\sin (x)} \frac{d}{d x} (\sin (x)) + \sin (x) \frac{d}{d x} (e^{\sin (x)})) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) \frac{d}{d x} (e^{\sin (x)})) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

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 $g (x) = \sin (x)$ .

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To apply the Chain Rule, set $u_{1}$ as $\sin (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

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

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) (e^{u_{1}} \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

Replace all occurrences of $u_{1}$ with $\sin (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) (e^{\sin (x)} \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + \frac{d}{d x} (e^{\sin (x)} \cos^{2} (x))$

Evaluate $\frac{d}{d x} [e^{\sin (x)} \cos^{2} (x)]$ .

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$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} \frac{d}{d x} (\cos^{2} (x)) + \cos^{2} (x) \frac{d}{d x} (e^{\sin (x)})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \cos (x)$ .

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To apply the Chain Rule, set $u_{2}$ as $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (\frac{d}{d u (2)} ({u_{2}}^{2}) \frac{d}{d x} (\cos (x))) + \cos^{2} (x) \frac{d}{d x} (e^{\sin (x)})$

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

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 u_{2} \frac{d}{d x} (\cos (x))) + \cos^{2} (x) \frac{d}{d x} (e^{\sin (x)})$

Replace all occurrences of $u_{2}$ with $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) \frac{d}{d x} (\cos (x))) + \cos^{2} (x) \frac{d}{d x} (e^{\sin (x)})$

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) (- \sin (x))) + \cos^{2} (x) \frac{d}{d x} (e^{\sin (x)})$

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 $g (x) = \sin (x)$ .

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To apply the Chain Rule, set $u_{3}$ as $\sin (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) (- \sin (x))) + \cos^{2} (x) (\frac{d}{d u (3)} (e^{u_{3}}) \frac{d}{d x} (\sin (x)))$

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

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) (- \sin (x))) + \cos^{2} (x) (e^{u_{3}} \frac{d}{d x} (\sin (x)))$

Replace all occurrences of $u_{3}$ with $\sin (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) (- \sin (x))) + \cos^{2} (x) (e^{\sin (x)} \frac{d}{d x} (\sin (x)))$

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (2 \cos (x) (- \sin (x))) + \cos^{2} (x) (e^{\sin (x)} \cos (x))$

Multiply $- 1$ by $2$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos^{2} (x) (e^{\sin (x)} \cos (x))$

Multiply $\cos^{2} (x)$ by $\cos (x)$ by adding the exponents.

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Move $\cos (x)$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos (x) \cos^{2} (x) e^{\sin (x)}$

Multiply $\cos (x)$ by $\cos^{2} (x)$ .

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Raise $\cos (x)$ to the power of $1$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos (x) \cos^{2} (x) e^{\sin (x)}$

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

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + {\cos (x)}^{1 + 2} e^{\sin (x)}$

Add $1$ and $2$ .

$f''' (x) = - (e^{\sin (x)} \cos (x) + \sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos^{3} (x) e^{\sin (x)}$

Simplify.

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Apply the distributive property.

$f''' (x) = - (e^{\sin (x)} \cos (x)) - (\sin (x) e^{\sin (x)} \cos (x)) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos^{3} (x) e^{\sin (x)}$

Combine terms.

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Reorder the factors of $- \sin (x) e^{\sin (x)} \cos (x)$ .

$f''' (x) = - e^{\sin (x)} \cos (x) - e^{\sin (x)} \cos (x) \sin (x) + e^{\sin (x)} (- 2 \cos (x) \sin (x)) + \cos^{3} (x) e^{\sin (x)}$

Add $- e^{\sin (x)} \cos (x) \sin (x)$ and $e^{\sin (x)} (- 2 \cos (x) \sin (x))$ .

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Reorder $e^{\sin (x)}$ and $- 2$ .

$f''' (x) = - e^{\sin (x)} \cos (x) - e^{\sin (x)} \cos (x) \sin (x) - 2 \cdot (e^{\sin (x)} \cos (x) \sin (x)) + \cos^{3} (x) e^{\sin (x)}$

Subtract $2 \cdot e^{\sin (x)} \cos (x) \sin (x)$ from $- e^{\sin (x)} \cos (x) \sin (x)$ .

$f''' (x) = - e^{\sin (x)} \cos (x) - 3 e^{\sin (x)} \cos (x) \sin (x) + \cos^{3} (x) e^{\sin (x)}$

Reorder terms.

$f''' (x) = - e^{\sin (x)} \cos (x) - 3 e^{\sin (x)} \cos (x) \sin (x) + e^{\sin (x)} \cos^{3} (x)$

Factor $\cos (x)$ out of $- e^{\sin (x)} \cos (x) - 3 e^{\sin (x)} \cos (x) \sin (x) + e^{\sin (x)} \cos^{3} (x)$ .

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Factor $\cos (x)$ out of $- e^{\sin (x)} \cos (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)}) - 3 e^{\sin (x)} \cos (x) \sin (x) + e^{\sin (x)} \cos^{3} (x)$

Factor $\cos (x)$ out of $- 3 e^{\sin (x)} \cos (x) \sin (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)}) + \cos (x) (- 3 e^{\sin (x)} \sin (x)) + e^{\sin (x)} \cos^{3} (x)$

Factor $\cos (x)$ out of $e^{\sin (x)} \cos^{3} (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)}) + \cos (x) (- 3 e^{\sin (x)} \sin (x)) + \cos (x) (e^{\sin (x)} \cos^{2} (x))$

Factor $\cos (x)$ out of $\cos (x) (- e^{\sin (x)}) + \cos (x) (- 3 e^{\sin (x)} \sin (x))$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} - 3 e^{\sin (x)} \sin (x)) + \cos (x) (e^{\sin (x)} \cos^{2} (x))$

Factor $\cos (x)$ out of $\cos (x) (- e^{\sin (x)} - 3 e^{\sin (x)} \sin (x)) + \cos (x) (e^{\sin (x)} \cos^{2} (x))$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} - 3 e^{\sin (x)} \sin (x) + e^{\sin (x)} \cos^{2} (x))$

Move $e^{\sin (x)} \cos^{2} (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} + e^{\sin (x)} \cos^{2} (x) - 3 e^{\sin (x)} \sin (x))$

Factor $- e^{\sin (x)}$ out of $- e^{\sin (x)}$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} \cdot 1 + e^{\sin (x)} \cos^{2} (x) - 3 e^{\sin (x)} \sin (x))$

Factor $- e^{\sin (x)}$ out of $e^{\sin (x)} \cos^{2} (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} \cdot 1 - e^{\sin (x)} (- 1 \cos^{2} (x)) - 3 e^{\sin (x)} \sin (x))$

Factor $- e^{\sin (x)}$ out of $- e^{\sin (x)} (1) - e^{\sin (x)} (- 1 \cos^{2} (x))$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} (1 - 1 \cos^{2} (x)) - 3 e^{\sin (x)} \sin (x))$

Rewrite $- 1 \cos^{2} (x)$ as $- \cos^{2} (x)$ .

$f''' (x) = \cos (x) (- e^{\sin (x)} (1 - \cos^{2} (x)) - 3 e^{\sin (x)} \sin (x))$

Apply pythagorean identity.

$f''' (x) = \cos (x) (- e^{\sin (x)} \sin^{2} (x) - 3 e^{\sin (x)} \sin (x))$

Apply the distributive property.

$f''' (x) = \cos (x) (- e^{\sin (x)} \sin^{2} (x)) + \cos (x) (- 3 e^{\sin (x)} \sin (x))$

Rewrite using the commutative property of multiplication.

$f''' (x) = - \cos (x) e^{\sin (x)} \sin^{2} (x) + \cos (x) (- 3 e^{\sin (x)} \sin (x))$

Rewrite using the commutative property of multiplication.

$f''' (x) = - \cos (x) e^{\sin (x)} \sin^{2} (x) - 3 \cos (x) e^{\sin (x)} \sin (x)$

Reorder factors in $- \cos (x) e^{\sin (x)} \sin^{2} (x) - 3 \cos (x) e^{\sin (x)} \sin (x)$ .

$f''' (x) = - e^{\sin (x)} \cos (x) \sin^{2} (x) - 3 e^{\sin (x)} \cos (x) \sin (x)$

The third derivative of $f (x)$ with respect to $x$ is $- e^{\sin (x)} \cos (x) \sin^{2} (x) - 3 e^{\sin (x)} \cos (x) \sin (x)$ .

$- e^{\sin (x)} \cos (x) \sin^{2} (x) - 3 e^{\sin (x)} \cos (x) \sin (x)$