Calculus Examples

$f (x) = \ln (x^{5} - e^{2 x})$

Step 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 $g (x) = x^{5} - e^{2 x}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $x^{5} - e^{2 x}$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [x^{5} - e^{2 x}]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [x^{5} - e^{2 x}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $x^{5} - e^{2 x}$ .

$\frac{1}{x^{5} - e^{2 x}} \frac{d}{d x} [x^{5} - e^{2 x}]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $x^{5} - e^{2 x}$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- e^{2 x}]$ .

$\frac{1}{x^{5} - e^{2 x}} (\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- e^{2 x}])$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} + \frac{d}{d x} [- e^{2 x}])$

Step 2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{2 x}$ with respect to $x$ is $- \frac{d}{d x} [e^{2 x}]$ .

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - \frac{d}{d x} [e^{2 x}])$

Step 3

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 $g (x) = 2 x$ .

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Step 3.1

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

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [2 x]))$

Step 3.2

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

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - (e^{u_{2}} \frac{d}{d x} [2 x]))$

Step 3.3

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

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - (e^{2 x} \frac{d}{d x} [2 x]))$

Step 4

Differentiate.

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Step 4.1

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

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - e^{2 x} (2 \frac{d}{d x} [x]))$

Step 4.2

Multiply $2$ by $- 1$ .

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - 2 e^{2 x} \frac{d}{d x} [x])$

Step 4.3

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

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - 2 e^{2 x} \cdot 1)$

Step 4.4

Multiply $- 2$ by $1$ .

$\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - 2 e^{2 x})$

Step 5

Simplify.

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Step 5.1

Reorder the factors of $\frac{1}{x^{5} - e^{2 x}} (5 x^{4} - 2 e^{2 x})$ .

$(5 x^{4} - 2 e^{2 x}) \frac{1}{x^{5} - e^{2 x}}$

Step 5.2

Multiply $5 x^{4} - 2 e^{2 x}$ by $\frac{1}{x^{5} - e^{2 x}}$ .

$\frac{5 x^{4} - 2 e^{2 x}}{x^{5} - e^{2 x}}$