Calculus Examples

$\int_{0}^{\ln (4 x)} \sin (2 e^{t}) d t$

Step 1

Take the derivative of $\int_{0}^{\ln (4 x)} \sin (2 e^{t}) d t$ with respect to $x$ using Fundamental Theorem of Calculus and the chain rule.

$\frac{d}{d x} [\ln (4 x)] \sin (2 e^{\ln (4 x)})$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 4 x$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $4 x$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [4 x] \sin (2 e^{\ln (4 x)})$

Step 2.2

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

$\frac{1}{u} \frac{d}{d x} [4 x] \sin (2 e^{\ln (4 x)})$

Step 2.3

Replace all occurrences of $u$ with $4 x$ .

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

Step 3

Differentiate.

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Step 3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

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

Step 3.2

Simplify terms.

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Step 3.2.1

Combine $4$ and $\frac{1}{4 x}$ .

$\frac{4}{4 x} \frac{d}{d x} [x] \sin (2 e^{\ln (4 x)})$

Step 3.2.2

Cancel the common factor of $4$ .

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Step 3.2.2.1

Cancel the common factor.

$\frac{4}{4 x} \frac{d}{d x} [x] \sin (2 e^{\ln (4 x)})$

Step 3.2.2.2

Rewrite the expression.

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

Step 3.3

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

$\frac{1}{x} \cdot 1 \sin (2 e^{\ln (4 x)})$

Step 3.4

Multiply $\frac{1}{x}$ by $1$ .

$\frac{1}{x} \sin (2 e^{\ln (4 x)})$

Step 4

Exponentiation and log are inverse functions.

$\frac{1}{x} \sin (2 (4 x))$

Step 5

Multiply $4$ by $2$ .

$\frac{1}{x} \sin (8 x)$

Step 6

Combine $\frac{1}{x}$ and $\sin (8 x)$ .

$\frac{\sin (8 x)}{x}$