Calculus Examples

$y = \sin (\ln (x)) + \cos (\ln (x))$

Step 1

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

$\frac{d}{d x} [\sin (\ln (x))] + \frac{d}{d x} [\cos (\ln (x))]$

Step 2

Evaluate $\frac{d}{d x} [\sin (\ln (x))]$ .

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$\cos (u_{1}) \frac{d}{d x} [\ln (x)] + \frac{d}{d x} [\cos (\ln (x))]$

Step 2.1.3

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

$\cos (\ln (x)) \frac{d}{d x} [\ln (x)] + \frac{d}{d x} [\cos (\ln (x))]$

Step 2.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\cos (\ln (x)) \frac{1}{x} + \frac{d}{d x} [\cos (\ln (x))]$

Step 2.3

Combine $\cos (\ln (x))$ and $\frac{1}{x}$ .

$\frac{\cos (\ln (x))}{x} + \frac{d}{d x} [\cos (\ln (x))]$

Step 3

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

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Step 3.1

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

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Step 3.1.1

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

$\frac{\cos (\ln (x))}{x} + \frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d x} [\ln (x)]$

Step 3.1.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 3.1.3

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

$\frac{\cos (\ln (x))}{x} - \sin (\ln (x)) \frac{d}{d x} [\ln (x)]$

Step 3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{\cos (\ln (x))}{x} - \sin (\ln (x)) \frac{1}{x}$

Step 3.3

Combine $\frac{1}{x}$ and $\sin (\ln (x))$ .

$\frac{\cos (\ln (x))}{x} - \frac{\sin (\ln (x))}{x}$