Calculus Examples

$y = {(3 x + 1)}^{x - 3}$

Step 1

Use the properties of logarithms to simplify the differentiation.

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Step 1.1

Rewrite ${(3 x + 1)}^{x - 3}$ as $e^{\ln ({(3 x + 1)}^{x - 3})}$ .

$\frac{d}{d x} [e^{\ln ({(3 x + 1)}^{x - 3})}]$

Step 1.2

Expand $\ln ({(3 x + 1)}^{x - 3})$ by moving $x - 3$ outside the logarithm.

$\frac{d}{d x} [e^{(x - 3) \ln (3 x + 1)}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = (x - 3) \ln (3 x + 1)$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $(x - 3) \ln (3 x + 1)$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [(x - 3) \ln (3 x + 1)]$

Step 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 $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [(x - 3) \ln (3 x + 1)]$

Step 2.3

Replace all occurrences of $u_{1}$ with $(x - 3) \ln (3 x + 1)$ .

$e^{(x - 3) \ln (3 x + 1)} \frac{d}{d x} [(x - 3) \ln (3 x + 1)]$

Step 3

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 $f (x) = x - 3$ and $g (x) = \ln (3 x + 1)$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{d}{d x} [\ln (3 x + 1)] + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 4

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 $g (x) = 3 x + 1$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $3 x + 1$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [3 x + 1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 4.2

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

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) (\frac{1}{u_{2}} \frac{d}{d x} [3 x + 1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 4.3

Replace all occurrences of $u_{2}$ with $3 x + 1$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) (\frac{1}{3 x + 1} \frac{d}{d x} [3 x + 1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $3 x + 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} (\frac{d}{d x} [3 x] + \frac{d}{d x} [1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.2

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

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} (3 \frac{d}{d x} [x] + \frac{d}{d x} [1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.3

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

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} (3 \cdot 1 + \frac{d}{d x} [1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.4

Multiply $3$ by $1$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} (3 + \frac{d}{d x} [1]) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} (3 + 0) + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.6

Combine fractions.

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Step 5.6.1

Add $3$ and $0$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{1}{3 x + 1} \cdot 3 + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.6.2

Combine $3$ and $\frac{1}{3 x + 1}$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1) \frac{d}{d x} [x - 3])$

Step 5.7

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1) (\frac{d}{d x} [x] + \frac{d}{d x} [- 3]))$

Step 5.8

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

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1) (1 + \frac{d}{d x} [- 3]))$

Step 5.9

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1) (1 + 0))$

Step 5.10

Simplify the expression.

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Step 5.10.1

Add $1$ and $0$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1) \cdot 1)$

Step 5.10.2

Multiply $\ln (3 x + 1)$ by $1$ .

$e^{(x - 3) \ln (3 x + 1)} ((x - 3) \frac{3}{3 x + 1} + \ln (3 x + 1))$