Calculus Examples

$y = 2^{x} \csc (2 x) + \ln (\cos (x))$

Step 1

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

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

Step 2

Evaluate $\frac{d}{d x} [2^{x} \csc (2 x)]$ .

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Step 2.1

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 $f (x) = 2^{x}$ and $g (x) = \csc (2 x)$ .

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

Step 2.2

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

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$2^{x} (\frac{d}{d u_{1}} [\csc (u_{1})] \frac{d}{d x} [2 x]) + \csc (2 x) \frac{d}{d x} [2^{x}] + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.2.2

The derivative of $\csc (u_{1})$ with respect to $u_{1}$ is $- \csc (u_{1}) \cot (u_{1})$ .

$2^{x} (- \csc (u_{1}) \cot (u_{1}) \frac{d}{d x} [2 x]) + \csc (2 x) \frac{d}{d x} [2^{x}] + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

$2^{x} (- \csc (2 x) \cot (2 x) \frac{d}{d x} [2 x]) + \csc (2 x) \frac{d}{d x} [2^{x}] + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.3

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

$2^{x} (- \csc (2 x) \cot (2 x) (2 \frac{d}{d x} [x])) + \csc (2 x) \frac{d}{d x} [2^{x}] + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.4

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

$2^{x} (- \csc (2 x) \cot (2 x) (2 \cdot 1)) + \csc (2 x) \frac{d}{d x} [2^{x}] + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.5

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

$2^{x} (- \csc (2 x) \cot (2 x) (2 \cdot 1)) + \csc (2 x) (2^{x} \ln (2)) + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.6

Multiply $2$ by $1$ .

$2^{x} (- \csc (2 x) \cot (2 x) \cdot 2) + \csc (2 x) (2^{x} \ln (2)) + \frac{d}{d x} [\ln (\cos (x))]$

Step 2.7

Multiply $2$ by $- 1$ .

$2^{x} (- 2 \csc (2 x) \cot (2 x)) + \csc (2 x) (2^{x} \ln (2)) + \frac{d}{d x} [\ln (\cos (x))]$

Step 3

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

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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 $f (x) = \ln (x)$ and $g (x) = \cos (x)$ .

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Step 3.1.1

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

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

Step 3.1.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$2^{x} (- 2 \csc (2 x) \cot (2 x)) + \csc (2 x) (2^{x} \ln (2)) + \frac{1}{u_{2}} \frac{d}{d x} [\cos (x)]$

Step 3.1.3

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

$2^{x} (- 2 \csc (2 x) \cot (2 x)) + \csc (2 x) (2^{x} \ln (2)) + \frac{1}{\cos (x)} \frac{d}{d x} [\cos (x)]$

Step 3.2

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

$2^{x} (- 2 \csc (2 x) \cot (2 x)) + \csc (2 x) (2^{x} \ln (2)) + \frac{1}{\cos (x)} (- \sin (x))$

Step 3.3

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$2^{x} (- 2 \csc (2 x) \cot (2 x)) + \csc (2 x) (2^{x} \ln (2)) + \sec (x) (- \sin (x))$

Step 4

Reorder terms.

$- 2 \cdot 2^{x} \cot (2 x) \csc (2 x) + 2^{x} \csc (2 x) \ln (2) - \sec (x) \sin (x)$