Calculus Examples

$y = 3 e^{4 x} + \log_{5} (7 x)$

Step 1

By the Sum Rule, the derivative of $3 e^{4 x} + \log_{5} (7 x)$ with respect to $x$ is $\frac{d}{d x} [3 e^{4 x}] + \frac{d}{d x} [\log_{5} (7 x)]$ .

$\frac{d}{d x} [3 e^{4 x}] + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2

Evaluate $\frac{d}{d x} [3 e^{4 x}]$ .

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Step 2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{4 x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{4 x}]$ .

$3 \frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [\log_{5} (7 x)]$

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 $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$3 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x]) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$3 (e^{u_{1}} \frac{d}{d x} [4 x]) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.2.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$3 (e^{4 x} \frac{d}{d x} [4 x]) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.3

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

$3 (e^{4 x} (4 \frac{d}{d x} [x])) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.4

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

$3 (e^{4 x} (4 \cdot 1)) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.5

Multiply $4$ by $1$ .

$3 (e^{4 x} \cdot 4) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.6

Move $4$ to the left of $e^{4 x}$ .

$3 (4 \cdot e^{4 x}) + \frac{d}{d x} [\log_{5} (7 x)]$

Step 2.7

Multiply $4$ by $3$ .

$12 e^{4 x} + \frac{d}{d x} [\log_{5} (7 x)]$

Step 3

Evaluate $\frac{d}{d x} [\log_{5} (7 x)]$ .

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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 $f (x) = \log_{5} (x)$ and $g (x) = 7 x$ .

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Step 3.1.1

To apply the Chain Rule, set $u_{2}$ as $7 x$ .

$12 e^{4 x} + \frac{d}{d u_{2}} [\log_{5} (u_{2})] \frac{d}{d x} [7 x]$

Step 3.1.2

The derivative of $\log_{5} (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2} \ln (5)}$ .

$12 e^{4 x} + \frac{1}{u_{2} \ln (5)} \frac{d}{d x} [7 x]$

Step 3.1.3

Replace all occurrences of $u_{2}$ with $7 x$ .

$12 e^{4 x} + \frac{1}{7 x \ln (5)} \frac{d}{d x} [7 x]$

Step 3.2

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$12 e^{4 x} + \frac{1}{7 x \ln (5)} (7 \frac{d}{d x} [x])$

Step 3.3

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

$12 e^{4 x} + \frac{1}{7 x \ln (5)} (7 \cdot 1)$

Step 3.4

Multiply $7$ by $1$ .

$12 e^{4 x} + \frac{1}{7 x \ln (5)} \cdot 7$

Step 3.5

Combine $\frac{1}{7 x \ln (5)}$ and $7$ .

$12 e^{4 x} + \frac{7}{7 x \ln (5)}$

Step 3.6

Cancel the common factor of $7$ .

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Step 3.6.1

Cancel the common factor.

$12 e^{4 x} + \frac{7}{7 x \ln (5)}$

Step 3.6.2

Rewrite the expression.

$12 e^{4 x} + \frac{1}{x \ln (5)}$

Step 4

Reorder terms.

$\frac{1}{x \ln (5)} + 12 e^{4 x}$