Calculus Examples

$y = \frac{{(3 x - 4)}^{4}}{{(2 x + 3)}^{7}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\frac{{(3 x - 4)}^{4}}{{(2 x + 3)}^{7}})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 $f (x) = {(3 x - 4)}^{4}$ and $g (x) = {(2 x + 3)}^{7}$ .

$\frac{{(2 x + 3)}^{7} \frac{d}{d x} [{(3 x - 4)}^{4}] - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{({(2 x + 3)}^{7})}^{2}}$

Step 3.2

Multiply the exponents in ${({(2 x + 3)}^{7})}^{2}$ .

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Step 3.2.1

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

$\frac{{(2 x + 3)}^{7} \frac{d}{d x} [{(3 x - 4)}^{4}] - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{7 \cdot 2}}$

Step 3.2.2

Multiply $7$ by $2$ .

$\frac{{(2 x + 3)}^{7} \frac{d}{d x} [{(3 x - 4)}^{4}] - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.3

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

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Step 3.3.1

To apply the Chain Rule, set $u_{1}$ as $3 x - 4$ .

$\frac{{(2 x + 3)}^{7} (\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d x} [3 x - 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 4$ .

$\frac{{(2 x + 3)}^{7} (4 {u_{1}}^{3} \frac{d}{d x} [3 x - 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.3.3

Replace all occurrences of $u_{1}$ with $3 x - 4$ .

$\frac{{(2 x + 3)}^{7} (4 {(3 x - 4)}^{3} \frac{d}{d x} [3 x - 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4

Differentiate.

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Step 3.4.1

Move $4$ to the left of ${(2 x + 3)}^{7}$ .

$\frac{4 \cdot {(2 x + 3)}^{7} {(3 x - 4)}^{3} \frac{d}{d x} [3 x - 4] - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.2

By the Sum Rule, the derivative of $3 x - 4$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 4]$ .

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} (\frac{d}{d x} [3 x] + \frac{d}{d x} [- 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.3

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

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} (3 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.4

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

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} (3 \cdot 1 + \frac{d}{d x} [- 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.5

Multiply $3$ by $1$ .

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} (3 + \frac{d}{d x} [- 4]) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.6

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} (3 + 0) - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.7

Simplify the expression.

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Step 3.4.7.1

Add $3$ and $0$ .

$\frac{4 {(2 x + 3)}^{7} {(3 x - 4)}^{3} \cdot 3 - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.4.7.2

Multiply $3$ by $4$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - {(3 x - 4)}^{4} \frac{d}{d x} [{(2 x + 3)}^{7}]}{{(2 x + 3)}^{14}}$

Step 3.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{7}$ and $g (x) = 2 x + 3$ .

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Step 3.5.1

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - {(3 x - 4)}^{4} (\frac{d}{d u_{2}} [{u_{2}}^{7}] \frac{d}{d x} [2 x + 3])}{{(2 x + 3)}^{14}}$

Step 3.5.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 7$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - {(3 x - 4)}^{4} (7 {u_{2}}^{6} \frac{d}{d x} [2 x + 3])}{{(2 x + 3)}^{14}}$

Step 3.5.3

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - {(3 x - 4)}^{4} (7 {(2 x + 3)}^{6} \frac{d}{d x} [2 x + 3])}{{(2 x + 3)}^{14}}$

Step 3.6

Differentiate.

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Step 3.6.1

Multiply $7$ by $- 1$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} \frac{d}{d x} [2 x + 3])}{{(2 x + 3)}^{14}}$

Step 3.6.2

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} (\frac{d}{d x} [2 x] + \frac{d}{d x} [3]))}{{(2 x + 3)}^{14}}$

Step 3.6.3

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} (2 \frac{d}{d x} [x] + \frac{d}{d x} [3]))}{{(2 x + 3)}^{14}}$

Step 3.6.4

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} (2 \cdot 1 + \frac{d}{d x} [3]))}{{(2 x + 3)}^{14}}$

Step 3.6.5

Multiply $2$ by $1$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} (2 + \frac{d}{d x} [3]))}{{(2 x + 3)}^{14}}$

Step 3.6.6

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

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} (2 + 0))}{{(2 x + 3)}^{14}}$

Step 3.6.7

Simplify the expression.

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Step 3.6.7.1

Add $2$ and $0$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} ({(2 x + 3)}^{6} \cdot 2)}{{(2 x + 3)}^{14}}$

Step 3.6.7.2

Move $2$ to the left of ${(2 x + 3)}^{6}$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 7 {(3 x - 4)}^{4} (2 \cdot {(2 x + 3)}^{6})}{{(2 x + 3)}^{14}}$

Step 3.6.7.3

Multiply $2$ by $- 7$ .

$\frac{12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 14 {(3 x - 4)}^{4} {(2 x + 3)}^{6}}{{(2 x + 3)}^{14}}$

Step 3.7

Simplify.

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Factor $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3}$ out of $12 {(2 x + 3)}^{7} {(3 x - 4)}^{3} - 14 {(3 x - 4)}^{4} {(2 x + 3)}^{6}$ .

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Step 3.7.1.1.1

Factor $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3}$ out of $12 {(2 x + 3)}^{7} {(3 x - 4)}^{3}$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (6 (2 x + 3)) - 14 {(3 x - 4)}^{4} {(2 x + 3)}^{6}}{{(2 x + 3)}^{14}}$

Step 3.7.1.1.2

Factor $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3}$ out of $- 14 {(3 x - 4)}^{4} {(2 x + 3)}^{6}$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (6 (2 x + 3)) + 2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (- 7 (3 x - 4))}{{(2 x + 3)}^{14}}$

Step 3.7.1.1.3

Factor $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3}$ out of $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (6 (2 x + 3)) + 2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (- 7 (3 x - 4))$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (6 (2 x + 3) - 7 (3 x - 4))}{{(2 x + 3)}^{14}}$

Step 3.7.1.2

Apply the distributive property.

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (6 (2 x) + 6 \cdot 3 - 7 (3 x - 4))}{{(2 x + 3)}^{14}}$

Step 3.7.1.3

Multiply $2$ by $6$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (12 x + 6 \cdot 3 - 7 (3 x - 4))}{{(2 x + 3)}^{14}}$

Step 3.7.1.4

Multiply $6$ by $3$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (12 x + 18 - 7 (3 x - 4))}{{(2 x + 3)}^{14}}$

Step 3.7.1.5

Apply the distributive property.

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (12 x + 18 - 7 (3 x) - 7 \cdot - 4)}{{(2 x + 3)}^{14}}$

Step 3.7.1.6

Multiply $3$ by $- 7$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (12 x + 18 - 21 x - 7 \cdot - 4)}{{(2 x + 3)}^{14}}$

Step 3.7.1.7

Multiply $- 7$ by $- 4$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (12 x + 18 - 21 x + 28)}{{(2 x + 3)}^{14}}$

Step 3.7.1.8

Subtract $21 x$ from $12 x$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (- 9 x + 18 + 28)}{{(2 x + 3)}^{14}}$

Step 3.7.1.9

Add $18$ and $28$ .

$\frac{2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (- 9 x + 46)}{{(2 x + 3)}^{14}}$

Step 3.7.2

Cancel the common factor of ${(2 x + 3)}^{6}$ and ${(2 x + 3)}^{14}$ .

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Step 3.7.2.1

Factor ${(2 x + 3)}^{6}$ out of $2 {(2 x + 3)}^{6} {(3 x - 4)}^{3} (- 9 x + 46)$ .

$\frac{{(2 x + 3)}^{6} (2 {(3 x - 4)}^{3} (- 9 x + 46))}{{(2 x + 3)}^{14}}$

Step 3.7.2.2

Cancel the common factors.

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Step 3.7.2.2.1

Factor ${(2 x + 3)}^{6}$ out of ${(2 x + 3)}^{14}$ .

$\frac{{(2 x + 3)}^{6} (2 {(3 x - 4)}^{3} (- 9 x + 46))}{{(2 x + 3)}^{6} {(2 x + 3)}^{8}}$

Step 3.7.2.2.2

Cancel the common factor.

$\frac{{(2 x + 3)}^{6} (2 {(3 x - 4)}^{3} (- 9 x + 46))}{{(2 x + 3)}^{6} {(2 x + 3)}^{8}}$

Step 3.7.2.2.3

Rewrite the expression.

$\frac{2 {(3 x - 4)}^{3} (- 9 x + 46)}{{(2 x + 3)}^{8}}$

Step 3.7.3

Factor $- 1$ out of $- 9 x$ .

$\frac{2 {(3 x - 4)}^{3} (- (9 x) + 46)}{{(2 x + 3)}^{8}}$

Step 3.7.4

Rewrite $46$ as $- 1 (- 46)$ .

$\frac{2 {(3 x - 4)}^{3} (- (9 x) - 1 (- 46))}{{(2 x + 3)}^{8}}$

Step 3.7.5

Factor $- 1$ out of $- (9 x) - 1 (- 46)$ .

$\frac{2 {(3 x - 4)}^{3} (- (9 x - 46))}{{(2 x + 3)}^{8}}$

Step 3.7.6

Rewrite $- (9 x - 46)$ as $- 1 (9 x - 46)$ .

$\frac{2 {(3 x - 4)}^{3} (- 1 (9 x - 46))}{{(2 x + 3)}^{8}}$

Step 3.7.7

Move the negative in front of the fraction.

$- \frac{(2 {(3 x - 4)}^{3}) (9 x - 46)}{{(2 x + 3)}^{8}}$

Step 3.7.8

Reorder factors in $- \frac{(2 {(3 x - 4)}^{3}) (9 x - 46)}{{(2 x + 3)}^{8}}$ .

$- \frac{2 {(3 x - 4)}^{3} (9 x - 46)}{{(2 x + 3)}^{8}}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = - \frac{2 {(3 x - 4)}^{3} (9 x - 46)}{{(2 x + 3)}^{8}}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{2 {(3 x - 4)}^{3} (9 x - 46)}{{(2 x + 3)}^{8}}$