Calculus Examples

$\frac{{(4 x - 2)}^{4}}{{(- 5 x + 1)}^{5}}$

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 $f (x) = {(4 x - 2)}^{4}$ and $g (x) = {(- 5 x + 1)}^{5}$ .

$\frac{{(- 5 x + 1)}^{5} \frac{d}{d x} [{(4 x - 2)}^{4}] - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{({(- 5 x + 1)}^{5})}^{2}}$

Step 2

Multiply the exponents in ${({(- 5 x + 1)}^{5})}^{2}$ .

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Step 2.1

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

$\frac{{(- 5 x + 1)}^{5} \frac{d}{d x} [{(4 x - 2)}^{4}] - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{5 \cdot 2}}$

Step 2.2

Multiply $5$ by $2$ .

$\frac{{(- 5 x + 1)}^{5} \frac{d}{d x} [{(4 x - 2)}^{4}] - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = 4 x - 2$ .

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Step 3.1

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

$\frac{{(- 5 x + 1)}^{5} (\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d x} [4 x - 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 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 $n = 4$ .

$\frac{{(- 5 x + 1)}^{5} (4 {u_{1}}^{3} \frac{d}{d x} [4 x - 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 3.3

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

$\frac{{(- 5 x + 1)}^{5} (4 {(4 x - 2)}^{3} \frac{d}{d x} [4 x - 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4

Differentiate.

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Step 4.1

Move $4$ to the left of ${(- 5 x + 1)}^{5}$ .

$\frac{4 \cdot {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} \frac{d}{d x} [4 x - 2] - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.2

By the Sum Rule, the derivative of $4 x - 2$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]$ .

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} (\frac{d}{d x} [4 x] + \frac{d}{d x} [- 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.3

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

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} (4 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.4

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

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} (4 \cdot 1 + \frac{d}{d x} [- 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.5

Multiply $4$ by $1$ .

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} (4 + \frac{d}{d x} [- 2]) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.6

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} (4 + 0) - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.7

Simplify the expression.

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Step 4.7.1

Add $4$ and $0$ .

$\frac{4 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} \cdot 4 - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 4.7.2

Multiply $4$ by $4$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - {(4 x - 2)}^{4} \frac{d}{d x} [{(- 5 x + 1)}^{5}]}{{(- 5 x + 1)}^{10}}$

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = - 5 x + 1$ .

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Step 5.1

To apply the Chain Rule, set $u_{2}$ as $- 5 x + 1$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - {(4 x - 2)}^{4} (\frac{d}{d u_{2}} [{u_{2}}^{5}] \frac{d}{d x} [- 5 x + 1])}{{(- 5 x + 1)}^{10}}$

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 $n = 5$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - {(4 x - 2)}^{4} (5 {u_{2}}^{4} \frac{d}{d x} [- 5 x + 1])}{{(- 5 x + 1)}^{10}}$

Step 5.3

Replace all occurrences of $u_{2}$ with $- 5 x + 1$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - {(4 x - 2)}^{4} (5 {(- 5 x + 1)}^{4} \frac{d}{d x} [- 5 x + 1])}{{(- 5 x + 1)}^{10}}$

Step 6

Differentiate.

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Step 6.1

Multiply $5$ by $- 1$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} \frac{d}{d x} [- 5 x + 1])}{{(- 5 x + 1)}^{10}}$

Step 6.2

By the Sum Rule, the derivative of $- 5 x + 1$ with respect to $x$ is $\frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} (\frac{d}{d x} [- 5 x] + \frac{d}{d x} [1]))}{{(- 5 x + 1)}^{10}}$

Step 6.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} (- 5 \frac{d}{d x} [x] + \frac{d}{d x} [1]))}{{(- 5 x + 1)}^{10}}$

Step 6.4

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

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} (- 5 \cdot 1 + \frac{d}{d x} [1]))}{{(- 5 x + 1)}^{10}}$

Step 6.5

Multiply $- 5$ by $1$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} (- 5 + \frac{d}{d x} [1]))}{{(- 5 x + 1)}^{10}}$

Step 6.6

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

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} (- 5 + 0))}{{(- 5 x + 1)}^{10}}$

Step 6.7

Simplify the expression.

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Step 6.7.1

Add $- 5$ and $0$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} ({(- 5 x + 1)}^{4} \cdot - 5)}{{(- 5 x + 1)}^{10}}$

Step 6.7.2

Move $- 5$ to the left of ${(- 5 x + 1)}^{4}$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} - 5 {(4 x - 2)}^{4} (- 5 \cdot {(- 5 x + 1)}^{4})}{{(- 5 x + 1)}^{10}}$

Step 6.7.3

Multiply $- 5$ by $- 5$ .

$\frac{16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} + 25 {(4 x - 2)}^{4} {(- 5 x + 1)}^{4}}{{(- 5 x + 1)}^{10}}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Factor ${(- 5 x + 1)}^{4} {(4 x - 2)}^{3}$ out of $16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3} + 25 {(4 x - 2)}^{4} {(- 5 x + 1)}^{4}$ .

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Step 7.1.1.1

Factor ${(- 5 x + 1)}^{4} {(4 x - 2)}^{3}$ out of $16 {(- 5 x + 1)}^{5} {(4 x - 2)}^{3}$ .

$\frac{{(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (16 (- 5 x + 1)) + 25 {(4 x - 2)}^{4} {(- 5 x + 1)}^{4}}{{(- 5 x + 1)}^{10}}$

Step 7.1.1.2

Factor ${(- 5 x + 1)}^{4} {(4 x - 2)}^{3}$ out of $25 {(4 x - 2)}^{4} {(- 5 x + 1)}^{4}$ .

$\frac{{(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (16 (- 5 x + 1)) + {(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.1.3

Factor ${(- 5 x + 1)}^{4} {(4 x - 2)}^{3}$ out of ${(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (16 (- 5 x + 1)) + {(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (25 (4 x - 2))$ .

$\frac{{(- 5 x + 1)}^{4} {(4 x - 2)}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.2

Factor $2$ out of $4 x - 2$ .

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Step 7.1.2.1

Factor $2$ out of $4 x$ .

$\frac{{(- 5 x + 1)}^{4} {(2 (2 x) - 2)}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.2.2

Factor $2$ out of $- 2$ .

$\frac{{(- 5 x + 1)}^{4} {(2 (2 x) + 2 (- 1))}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.2.3

Factor $2$ out of $2 (2 x) + 2 (- 1)$ .

$\frac{{(- 5 x + 1)}^{4} {(2 (2 x - 1))}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.3

Apply the product rule to $2 (2 x - 1)$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 2^{3} {(2 x - 1)}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.4

Raise $2$ to the power of $3$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (16 (- 5 x + 1) + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.5

Simplify each term.

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Step 7.1.5.1

Apply the distributive property.

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (16 (- 5 x) + 16 \cdot 1 + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.5.2

Multiply $- 5$ by $16$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (- 80 x + 16 \cdot 1 + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.5.3

Multiply $16$ by $1$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (- 80 x + 16 + 25 (4 x - 2))}{{(- 5 x + 1)}^{10}}$

Step 7.1.5.4

Apply the distributive property.

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (- 80 x + 16 + 25 (4 x) + 25 \cdot - 2)}{{(- 5 x + 1)}^{10}}$

Step 7.1.5.5

Multiply $4$ by $25$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (- 80 x + 16 + 100 x + 25 \cdot - 2)}{{(- 5 x + 1)}^{10}}$

Step 7.1.5.6

Multiply $25$ by $- 2$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (- 80 x + 16 + 100 x - 50)}{{(- 5 x + 1)}^{10}}$

Step 7.1.6

Add $- 80 x$ and $100 x$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (20 x + 16 - 50)}{{(- 5 x + 1)}^{10}}$

Step 7.1.7

Subtract $50$ from $16$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (20 x - 34)}{{(- 5 x + 1)}^{10}}$

Step 7.1.8

Factor $2$ out of $20 x - 34$ .

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Step 7.1.8.1

Factor $2$ out of $20 x$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (2 (10 x) - 34)}{{(- 5 x + 1)}^{10}}$

Step 7.1.8.2

Factor $2$ out of $- 34$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (2 (10 x) + 2 (- 17))}{{(- 5 x + 1)}^{10}}$

Step 7.1.8.3

Factor $2$ out of $2 (10 x) + 2 (- 17)$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} (2 (10 x - 17))}{{(- 5 x + 1)}^{10}}$

$\frac{{(- 5 x + 1)}^{4} \cdot 8 {(2 x - 1)}^{3} \cdot 2 (10 x - 17)}{{(- 5 x + 1)}^{10}}$

Step 7.1.9

Multiply $2$ by $8$ .

$\frac{{(- 5 x + 1)}^{4} \cdot 16 {(2 x - 1)}^{3} (10 x - 17)}{{(- 5 x + 1)}^{10}}$

Step 7.2

Combine terms.

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Step 7.2.1

Move $16$ to the left of ${(- 5 x + 1)}^{4}$ .

$\frac{16 \cdot {(- 5 x + 1)}^{4} {(2 x - 1)}^{3} (10 x - 17)}{{(- 5 x + 1)}^{10}}$

Step 7.2.2

Cancel the common factor of ${(- 5 x + 1)}^{4}$ and ${(- 5 x + 1)}^{10}$ .

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Step 7.2.2.1

Factor ${(- 5 x + 1)}^{4}$ out of $16 {(- 5 x + 1)}^{4} {(2 x - 1)}^{3} (10 x - 17)$ .

$\frac{{(- 5 x + 1)}^{4} (16 {(2 x - 1)}^{3} (10 x - 17))}{{(- 5 x + 1)}^{10}}$

Step 7.2.2.2

Cancel the common factors.

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Step 7.2.2.2.1

Factor ${(- 5 x + 1)}^{4}$ out of ${(- 5 x + 1)}^{10}$ .

$\frac{{(- 5 x + 1)}^{4} (16 {(2 x - 1)}^{3} (10 x - 17))}{{(- 5 x + 1)}^{4} {(- 5 x + 1)}^{6}}$

Step 7.2.2.2.2

Cancel the common factor.

$\frac{{(- 5 x + 1)}^{4} (16 {(2 x - 1)}^{3} (10 x - 17))}{{(- 5 x + 1)}^{4} {(- 5 x + 1)}^{6}}$

Step 7.2.2.2.3

Rewrite the expression.

$\frac{16 {(2 x - 1)}^{3} (10 x - 17)}{{(- 5 x + 1)}^{6}}$