Calculus Examples

$y = \frac{\cos (x)}{e^{x}}$

Step 1

Find the first derivative.

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Step 1.1

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

$\frac{e^{x} \frac{d}{d x} [\cos (x)] - \cos (x) \frac{d}{d x} [e^{x}]}{{(e^{x})}^{2}}$

Step 1.2

Multiply the exponents in ${(e^{x})}^{2}$ .

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Step 1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{e^{x} \frac{d}{d x} [\cos (x)] - \cos (x) \frac{d}{d x} [e^{x}]}{e^{x \cdot 2}}$

Step 1.2.2

Move $2$ to the left of $x$ .

$\frac{e^{x} \frac{d}{d x} [\cos (x)] - \cos (x) \frac{d}{d x} [e^{x}]}{e^{2 x}}$

Step 1.3

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

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

Step 1.4

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

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

Step 1.5

Simplify.

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Step 1.5.1

Simplify the numerator.

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Step 1.5.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.5.1.2

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

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

Step 1.5.2

Simplify the numerator.

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Step 1.5.2.1

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

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Step 1.5.2.1.1

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

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

Step 1.5.2.1.2

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

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

Step 1.5.2.1.3

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

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

Step 1.5.2.2

Factor $- 1$ out of $- \sin (x) - \cos (x)$ .

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Step 1.5.2.2.1

Factor $- 1$ out of $- \sin (x)$ .

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

Step 1.5.2.2.2

Factor $- 1$ out of $- \cos (x)$ .

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

Step 1.5.2.2.3

Factor $- 1$ out of $- (\sin (x)) - (\cos (x))$ .

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

$\frac{e^{x} \cdot - 1 (\sin (x) + \cos (x))}{e^{2 x}}$

Step 1.5.2.3

Factor out negative.

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

Step 1.5.3

Cancel the common factor of $e^{x}$ and $e^{2 x}$ .

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Step 1.5.3.1

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

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

Step 1.5.3.2

Cancel the common factors.

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Step 1.5.3.2.1

Multiply by $1$ .

$\frac{e^{2 x} (- e^{- x} (\sin (x) + \cos (x)))}{e^{2 x} \cdot 1}$

Step 1.5.3.2.2

Cancel the common factor.

$\frac{e^{2 x} (- e^{- x} (\sin (x) + \cos (x)))}{e^{2 x} \cdot 1}$

Step 1.5.3.2.3

Rewrite the expression.

$\frac{- e^{- x} (\sin (x) + \cos (x))}{1}$

Step 1.5.3.2.4

Divide $- e^{- x} (\sin (x) + \cos (x))$ by $1$ .

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

Step 1.5.4

Apply the distributive property.

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

Step 2

Find the second derivative.

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Step 2.1

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

$\frac{d}{d x} [- e^{- x} \sin (x)] + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2

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

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Step 2.2.1

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

$- \frac{d}{d x} [e^{- x} \sin (x)] + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.2

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

$- (e^{- x} \frac{d}{d x} [\sin (x)] + \sin (x) \frac{d}{d x} [e^{- x}]) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.3

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

$- (e^{- x} \cos (x) + \sin (x) \frac{d}{d x} [e^{- x}]) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.4

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 $g (x) = - x$ .

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Step 2.2.4.1

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

$- (e^{- x} \cos (x) + \sin (x) (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- x])) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.4.2

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

$- (e^{- x} \cos (x) + \sin (x) (e^{u_{1}} \frac{d}{d x} [- x])) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.4.3

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

$- (e^{- x} \cos (x) + \sin (x) (e^{- x} \frac{d}{d x} [- x])) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.5

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

$- (e^{- x} \cos (x) + \sin (x) (e^{- x} (- \frac{d}{d x} [x]))) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.6

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

$- (e^{- x} \cos (x) + \sin (x) (e^{- x} (- 1 \cdot 1))) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.7

Multiply $- 1$ by $1$ .

$- (e^{- x} \cos (x) + \sin (x) (e^{- x} \cdot - 1)) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.8

Move $- 1$ to the left of $e^{- x}$ .

$- (e^{- x} \cos (x) + \sin (x) (- 1 \cdot e^{- x})) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.2.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) + \frac{d}{d x} [- e^{- x} \cos (x)]$

Step 2.3

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

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Step 2.3.1

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - \frac{d}{d x} [e^{- x} \cos (x)]$

Step 2.3.2

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} \frac{d}{d x} [\cos (x)] + \cos (x) \frac{d}{d x} [e^{- x}])$

Step 2.3.3

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) \frac{d}{d x} [e^{- x}])$

Step 2.3.4

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 $g (x) = - x$ .

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Step 2.3.4.1

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]))$

Step 2.3.4.2

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (e^{u_{2}} \frac{d}{d x} [- x]))$

Step 2.3.4.3

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} \frac{d}{d x} [- x]))$

Step 2.3.5

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} (- \frac{d}{d x} [x])))$

Step 2.3.6

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

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} (- 1 \cdot 1)))$

Step 2.3.7

Multiply $- 1$ by $1$ .

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} \cdot - 1))$

Step 2.3.8

Move $- 1$ to the left of $e^{- x}$ .

$- (e^{- x} \cos (x) + \sin (x) (- e^{- x})) - (e^{- x} (- \sin (x)) + \cos (x) (- 1 \cdot e^{- x}))$

Step 2.3.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Apply the distributive property.

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

Step 2.4.3

Combine terms.

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Step 2.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.4.3.2

Multiply $\sin (x)$ by $1$ .

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

Step 2.4.3.3

Multiply $- 1$ by $- 1$ .

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

Step 2.4.3.4

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

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

Step 2.4.3.5

Multiply $- 1$ by $- 1$ .

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

Step 2.4.3.6

Multiply $\cos (x)$ by $1$ .

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

Step 2.4.3.7

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

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Step 2.4.3.7.1

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

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

Step 2.4.3.7.2

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

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

Step 2.4.3.8

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

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Step 2.4.3.8.1

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

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

Step 2.4.3.8.2

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

$2 e^{- x} \sin (x) + 0$

Step 2.4.3.9

Add $2 e^{- x} \sin (x)$ and $0$ .

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