Calculus Examples

$y = \frac{e^{2 x + 3}}{{(4 x - 1)}^{3}}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = e^{2 x + 3}$ and $g (x) = {(4 x - 1)}^{3}$ .

$\frac{{(4 x - 1)}^{3} \frac{d}{d x} [e^{2 x + 3}] - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{({(4 x - 1)}^{3})}^{2}}$

Step 2

Multiply the exponents in ${({(4 x - 1)}^{3})}^{2}$ .

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Step 2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{{(4 x - 1)}^{3} \frac{d}{d x} [e^{2 x + 3}] - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{3 \cdot 2}}$

Step 2.2

Multiply $3$ by $2$ .

$\frac{{(4 x - 1)}^{3} \frac{d}{d x} [e^{2 x + 3}] - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

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 $g (x) = 2 x + 3$ .

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Step 3.1

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

$\frac{{(4 x - 1)}^{3} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x + 3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 3.2

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

$\frac{{(4 x - 1)}^{3} (e^{u_{1}} \frac{d}{d x} [2 x + 3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 3.3

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

$\frac{{(4 x - 1)}^{3} (e^{2 x + 3} \frac{d}{d x} [2 x + 3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4

Differentiate.

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Step 4.1

By the Sum Rule, the derivative of $2 x + 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [3]$ .

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} (\frac{d}{d x} [2 x] + \frac{d}{d x} [3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.2

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

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} (2 \frac{d}{d x} [x] + \frac{d}{d x} [3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.3

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

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} (2 \cdot 1 + \frac{d}{d x} [3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.4

Multiply $2$ by $1$ .

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} (2 + \frac{d}{d x} [3]) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.5

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

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} (2 + 0) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.6

Simplify the expression.

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Step 4.6.1

Add $2$ and $0$ .

$\frac{{(4 x - 1)}^{3} e^{2 x + 3} \cdot 2 - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 4.6.2

Move $2$ to the left of ${(4 x - 1)}^{3} e^{2 x + 3}$ .

$\frac{2 \cdot ({(4 x - 1)}^{3} e^{2 x + 3}) - e^{2 x + 3} \frac{d}{d x} [{(4 x - 1)}^{3}]}{{(4 x - 1)}^{6}}$

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = 4 x - 1$ .

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Step 5.1

To apply the Chain Rule, set $u_{2}$ as $4 x - 1$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - e^{2 x + 3} (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [4 x - 1])}{{(4 x - 1)}^{6}}$

Step 5.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - e^{2 x + 3} (3 {u_{2}}^{2} \frac{d}{d x} [4 x - 1])}{{(4 x - 1)}^{6}}$

Step 5.3

Replace all occurrences of $u_{2}$ with $4 x - 1$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - e^{2 x + 3} (3 {(4 x - 1)}^{2} \frac{d}{d x} [4 x - 1])}{{(4 x - 1)}^{6}}$

Step 6

Differentiate.

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Step 6.1

Multiply $3$ by $- 1$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} \frac{d}{d x} [4 x - 1])}{{(4 x - 1)}^{6}}$

Step 6.2

By the Sum Rule, the derivative of $4 x - 1$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} (\frac{d}{d x} [4 x] + \frac{d}{d x} [- 1]))}{{(4 x - 1)}^{6}}$

Step 6.3

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

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} (4 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]))}{{(4 x - 1)}^{6}}$

Step 6.4

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

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} (4 \cdot 1 + \frac{d}{d x} [- 1]))}{{(4 x - 1)}^{6}}$

Step 6.5

Multiply $4$ by $1$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} (4 + \frac{d}{d x} [- 1]))}{{(4 x - 1)}^{6}}$

Step 6.6

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

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} (4 + 0))}{{(4 x - 1)}^{6}}$

Step 6.7

Simplify the expression.

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Step 6.7.1

Add $4$ and $0$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} ({(4 x - 1)}^{2} \cdot 4)}{{(4 x - 1)}^{6}}$

Step 6.7.2

Move $4$ to the left of ${(4 x - 1)}^{2}$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 3 e^{2 x + 3} (4 \cdot {(4 x - 1)}^{2})}{{(4 x - 1)}^{6}}$

Step 6.7.3

Multiply $4$ by $- 3$ .

$\frac{2 {(4 x - 1)}^{3} e^{2 x + 3} - 12 e^{2 x + 3} {(4 x - 1)}^{2}}{{(4 x - 1)}^{6}}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Factor $2 {(4 x - 1)}^{2} e^{2 x + 3}$ out of $2 {(4 x - 1)}^{3} e^{2 x + 3} - 12 e^{2 x + 3} {(4 x - 1)}^{2}$ .

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Step 7.1.1.1

Factor $2 {(4 x - 1)}^{2} e^{2 x + 3}$ out of $2 {(4 x - 1)}^{3} e^{2 x + 3}$ .

$\frac{2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 1) - 12 e^{2 x + 3} {(4 x - 1)}^{2}}{{(4 x - 1)}^{6}}$

Step 7.1.1.2

Factor $2 {(4 x - 1)}^{2} e^{2 x + 3}$ out of $- 12 e^{2 x + 3} {(4 x - 1)}^{2}$ .

$\frac{2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 1) + 2 {(4 x - 1)}^{2} e^{2 x + 3} (- 6)}{{(4 x - 1)}^{6}}$

Step 7.1.1.3

Factor $2 {(4 x - 1)}^{2} e^{2 x + 3}$ out of $2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 1) + 2 {(4 x - 1)}^{2} e^{2 x + 3} (- 6)$ .

$\frac{2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 1 - 6)}{{(4 x - 1)}^{6}}$

Step 7.1.2

Subtract $6$ from $- 1$ .

$\frac{2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 7)}{{(4 x - 1)}^{6}}$

Step 7.2

Cancel the common factor of ${(4 x - 1)}^{2}$ and ${(4 x - 1)}^{6}$ .

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Step 7.2.1

Factor ${(4 x - 1)}^{2}$ out of $2 {(4 x - 1)}^{2} e^{2 x + 3} (4 x - 7)$ .

$\frac{{(4 x - 1)}^{2} (2 e^{2 x + 3} (4 x - 7))}{{(4 x - 1)}^{6}}$

Step 7.2.2

Cancel the common factors.

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Step 7.2.2.1

Factor ${(4 x - 1)}^{2}$ out of ${(4 x - 1)}^{6}$ .

$\frac{{(4 x - 1)}^{2} (2 e^{2 x + 3} (4 x - 7))}{{(4 x - 1)}^{2} {(4 x - 1)}^{4}}$

Step 7.2.2.2

Cancel the common factor.

$\frac{{(4 x - 1)}^{2} (2 e^{2 x + 3} (4 x - 7))}{{(4 x - 1)}^{2} {(4 x - 1)}^{4}}$

Step 7.2.2.3

Rewrite the expression.

$\frac{2 e^{2 x + 3} (4 x - 7)}{{(4 x - 1)}^{4}}$