Calculus Examples

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

Find the first 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 $f (x) = e^{x}$ and $g (x) = \cos (x)$ .

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

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

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

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

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

Reorder terms.

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

Find the second derivative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Reorder $e^{x}$ and $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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