Calculus Examples

$x {(y + 2)}^{5} = 9$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x {(y + 2)}^{5}) = \frac{d}{d x} (9)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Use the Binomial Theorem.

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

Step 2.2

Simplify each term.

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Step 2.2.1

Multiply $2$ by $5$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $4$ by $10$ .

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 10 y^{2} \cdot 2^{3} + 5 y \cdot 2^{4} + 2^{5})]$

Step 2.2.4

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

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 10 y^{2} \cdot 8 + 5 y \cdot 2^{4} + 2^{5})]$

Step 2.2.5

Multiply $8$ by $10$ .

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 5 y \cdot 2^{4} + 2^{5})]$

Step 2.2.6

Raise $2$ to the power of $4$ .

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 5 y \cdot 16 + 2^{5})]$

Step 2.2.7

Multiply $16$ by $5$ .

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 2^{5})]$

Step 2.2.8

Raise $2$ to the power of $5$ .

$\frac{d}{d x} [x (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32)]$

Step 2.3

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 $f (x) = x$ and $g (x) = y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$ .

$x \frac{d}{d x} [y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32] + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.4

By the Sum Rule, the derivative of $y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$ with respect to $x$ is $\frac{d}{d x} [y^{5}] + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]$ .

$x (\frac{d}{d x} [y^{5}] + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.5

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 $g (x) = y$ .

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Step 2.5.1

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

$x (\frac{d}{d u_{1}} [{u_{1}}^{5}] \frac{d}{d x} [y] + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.5.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 5$ .

$x (5 {u_{1}}^{4} \frac{d}{d x} [y] + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.5.3

Replace all occurrences of $u_{1}$ with $y$ .

$x (5 y^{4} \frac{d}{d x} [y] + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.6

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (5 y^{4} y' + \frac{d}{d x} [10 y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.7

Since $10$ is constant with respect to $x$ , the derivative of $10 y^{4}$ with respect to $x$ is $10 \frac{d}{d x} [y^{4}]$ .

$x (5 y^{4} y' + 10 \frac{d}{d x} [y^{4}] + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.8

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 $g (x) = y$ .

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Step 2.8.1

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

$x (5 y^{4} y' + 10 (\frac{d}{d u_{2}} [{u_{2}}^{4}] \frac{d}{d x} [y]) + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.8.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 4$ .

$x (5 y^{4} y' + 10 (4 {u_{2}}^{3} \frac{d}{d x} [y]) + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.8.3

Replace all occurrences of $u_{2}$ with $y$ .

$x (5 y^{4} y' + 10 (4 y^{3} \frac{d}{d x} [y]) + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.9

Multiply $4$ by $10$ .

$x (5 y^{4} y' + 40 (y^{3} \frac{d}{d x} [y]) + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.10

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (5 y^{4} y' + 40 y^{3} y' + \frac{d}{d x} [40 y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.11

Since $40$ is constant with respect to $x$ , the derivative of $40 y^{3}$ with respect to $x$ is $40 \frac{d}{d x} [y^{3}]$ .

$x (5 y^{4} y' + 40 y^{3} y' + 40 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.12

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 $g (x) = y$ .

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Step 2.12.1

To apply the Chain Rule, set $u_{3}$ as $y$ .

$x (5 y^{4} y' + 40 y^{3} y' + 40 (\frac{d}{d u_{3}} [{u_{3}}^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.12.2

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

$x (5 y^{4} y' + 40 y^{3} y' + 40 (3 {u_{3}}^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.12.3

Replace all occurrences of $u_{3}$ with $y$ .

$x (5 y^{4} y' + 40 y^{3} y' + 40 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.13

Multiply $3$ by $40$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 (y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.14

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + \frac{d}{d x} [80 y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.15

Since $80$ is constant with respect to $x$ , the derivative of $80 y^{2}$ with respect to $x$ is $80 \frac{d}{d x} [y^{2}]$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 80 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.16

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

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Step 2.16.1

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

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 80 (\frac{d}{d u_{4}} [{u_{4}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.16.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{4}} [{u_{4}}^{n}]$ is $n {u_{4}}^{n - 1}$ where $n = 2$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 80 (2 u_{4} \frac{d}{d x} [y]) + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.16.3

Replace all occurrences of $u_{4}$ with $y$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 80 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.17

Multiply $2$ by $80$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 (y \frac{d}{d x} [y]) + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.18

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + \frac{d}{d x} [80 y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.19

Since $80$ is constant with respect to $x$ , the derivative of $80 y$ with respect to $x$ is $80 \frac{d}{d x} [y]$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 \frac{d}{d x} [y] + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.20

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y' + \frac{d}{d x} [32]) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.21

Since $32$ is constant with respect to $x$ , the derivative of $32$ with respect to $x$ is $0$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y' + 0) + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.22

Add $5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y'$ and $0$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y') + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \frac{d}{d x} [x]$

Step 2.23

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

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y') + (y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32) \cdot 1$

Step 2.24

Multiply $y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$ by $1$ .

$x (5 y^{4} y' + 40 y^{3} y' + 120 y^{2} y' + 160 y y' + 80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25

Simplify.

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Step 2.25.1

Apply the distributive property.

$x (5 y^{4} y') + x (40 y^{3} y') + x (120 y^{2} y') + x (160 y y') + x (80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.2

Combine terms.

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Step 2.25.2.1

Move $5$ to the left of $x$ .

$5 \cdot x y^{4} y' + x (40 y^{3} y') + x (120 y^{2} y') + x (160 y y') + x (80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.2.2

Move $40$ to the left of $x$ .

$5 x y^{4} y' + 40 \cdot x y^{3} y' + x (120 y^{2} y') + x (160 y y') + x (80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.2.3

Move $120$ to the left of $x$ .

$5 x y^{4} y' + 40 x y^{3} y' + 120 \cdot x y^{2} y' + x (160 y y') + x (80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.2.4

Move $160$ to the left of $x$ .

$5 x y^{4} y' + 40 x y^{3} y' + 120 x y^{2} y' + 160 \cdot x y y' + x (80 y') + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.2.5

Move $80$ to the left of $x$ .

$5 x y^{4} y' + 40 x y^{3} y' + 120 x y^{2} y' + 160 x y y' + 80 x y' + y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 80 y + 32$

Step 2.25.3

Reorder terms.

$y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32$

Step 3

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$0$

Step 4

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

$y^{5} + 10 y^{4} + 40 y^{3} + 80 y^{2} + 5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32 = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.1.1

Subtract $y^{5}$ from both sides of the equation.

$10 y^{4} + 40 y^{3} + 80 y^{2} + 5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32 = - y^{5}$

Step 5.1.2

Subtract $10 y^{4}$ from both sides of the equation.

$40 y^{3} + 80 y^{2} + 5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32 = - y^{5} - 10 y^{4}$

Step 5.1.3

Subtract $40 y^{3}$ from both sides of the equation.

$80 y^{2} + 5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32 = - y^{5} - 10 y^{4} - 40 y^{3}$

Step 5.1.4

Subtract $80 y^{2}$ from both sides of the equation.

$5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 80 y + 32 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2}$

Step 5.1.5

Subtract $80 y$ from both sides of the equation.

$5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' + 32 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y$

Step 5.1.6

Subtract $32$ from both sides of the equation.

$5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2

Factor $5 x y'$ out of $5 y^{4} x y' + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y'$ .

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Step 5.2.1

Factor $5 x y'$ out of $5 y^{4} x y'$ .

$5 x y' y^{4} + 40 y^{3} x y' + 120 y^{2} x y' + 160 x y y' + 80 x y' = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.2

Factor $5 x y'$ out of $40 y^{3} x y'$ .

$5 x y' y^{4} + 5 x y' (8 y^{3}) + 120 y^{2} x y' + 160 x y y' + 80 x y' = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.3

Factor $5 x y'$ out of $120 y^{2} x y'$ .

$5 x y' y^{4} + 5 x y' (8 y^{3}) + 5 x y' (24 y^{2}) + 160 x y y' + 80 x y' = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.4

Factor $5 x y'$ out of $160 x y y'$ .

$5 x y' y^{4} + 5 x y' (8 y^{3}) + 5 x y' (24 y^{2}) + 5 x y' (32 y) + 80 x y' = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.5

Factor $5 x y'$ out of $80 x y'$ .

$5 x y' y^{4} + 5 x y' (8 y^{3}) + 5 x y' (24 y^{2}) + 5 x y' (32 y) + 5 x y' \cdot 16 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.6

Factor $5 x y'$ out of $5 x y' y^{4} + 5 x y' (8 y^{3})$ .

$5 x y' (y^{4} + 8 y^{3}) + 5 x y' (24 y^{2}) + 5 x y' (32 y) + 5 x y' \cdot 16 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.7

Factor $5 x y'$ out of $5 x y' (y^{4} + 8 y^{3}) + 5 x y' (24 y^{2})$ .

$5 x y' (y^{4} + 8 y^{3} + 24 y^{2}) + 5 x y' (32 y) + 5 x y' \cdot 16 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.8

Factor $5 x y'$ out of $5 x y' (y^{4} + 8 y^{3} + 24 y^{2}) + 5 x y' (32 y)$ .

$5 x y' (y^{4} + 8 y^{3} + 24 y^{2} + 32 y) + 5 x y' \cdot 16 = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.2.9

Factor $5 x y'$ out of $5 x y' (y^{4} + 8 y^{3} + 24 y^{2} + 32 y) + 5 x y' \cdot 16$ .

$5 x y' (y^{4} + 8 y^{3} + 24 y^{2} + 32 y + 16) = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.3

Make each term match the terms from the binomial theorem formula.

$5 x y' (y^{4} + 4 \cdot (2 y^{3}) + 6 \cdot (2^{2} y^{2}) + 4 \cdot (2^{3} y) + 2^{4}) = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.4

Factor $y^{4} + 4 \cdot 2 y^{3} + 6 \cdot 2^{2} y^{2} + 4 \cdot 2^{3} y + 2^{4}$ using the binomial theorem.

$5 x y' {(y + 2)}^{4} = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$

Step 5.5

Divide each term in $5 x y' {(y + 2)}^{4} = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$ by $5 x {(y + 2)}^{4}$ and simplify.

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Step 5.5.1

Divide each term in $5 x y' {(y + 2)}^{4} = - y^{5} - 10 y^{4} - 40 y^{3} - 80 y^{2} - 80 y - 32$ by $5 x {(y + 2)}^{4}$ .

$\frac{5 x y' {(y + 2)}^{4}}{5 x {(y + 2)}^{4}} = \frac{- y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 10 y^{4}}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $5$ .

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Step 5.5.2.1.1

Cancel the common factor.

Step 5.5.2.1.2

Rewrite the expression.

$\frac{x y' {(y + 2)}^{4}}{x {(y + 2)}^{4}} = \frac{- y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 10 y^{4}}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.2.2

Cancel the common factor of $x$ .

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Step 5.5.2.2.1

Cancel the common factor.

Step 5.5.2.2.2

Rewrite the expression.

$\frac{y' {(y + 2)}^{4}}{{(y + 2)}^{4}} = \frac{- y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 10 y^{4}}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.2.3

Cancel the common factor of ${(y + 2)}^{4}$ .

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Step 5.5.2.3.1

Cancel the common factor.

Step 5.5.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{- y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 10 y^{4}}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Simplify each term.

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Step 5.5.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 10 y^{4}}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.2

Cancel the common factor of $- 10$ and $5$ .

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Step 5.5.3.1.2.1

Factor $5$ out of $- 10 y^{4}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} + \frac{5 (- 2 y^{4})}{5 x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.2.2

Cancel the common factors.

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Step 5.5.3.1.2.2.1

Factor $5$ out of $5 x {(y + 2)}^{4}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} + \frac{5 (- 2 y^{4})}{5 (x {(y + 2)}^{4})} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.2.2.2

Cancel the common factor.

Step 5.5.3.1.2.2.3

Rewrite the expression.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} + \frac{- 2 y^{4}}{x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.3

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} + \frac{- 40 y^{3}}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.4

Cancel the common factor of $- 40$ and $5$ .

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Step 5.5.3.1.4.1

Factor $5$ out of $- 40 y^{3}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} + \frac{5 (- 8 y^{3})}{5 x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.4.2

Cancel the common factors.

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Step 5.5.3.1.4.2.1

Factor $5$ out of $5 x {(y + 2)}^{4}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} + \frac{5 (- 8 y^{3})}{5 (x {(y + 2)}^{4})} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.4.2.2

Cancel the common factor.

Step 5.5.3.1.4.2.3

Rewrite the expression.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} + \frac{- 8 y^{3}}{x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.5

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} + \frac{- 80 y^{2}}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.6

Cancel the common factor of $- 80$ and $5$ .

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Step 5.5.3.1.6.1

Factor $5$ out of $- 80 y^{2}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} + \frac{5 (- 16 y^{2})}{5 x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.6.2

Cancel the common factors.

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Step 5.5.3.1.6.2.1

Factor $5$ out of $5 x {(y + 2)}^{4}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} + \frac{5 (- 16 y^{2})}{5 (x {(y + 2)}^{4})} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.6.2.2

Cancel the common factor.

Step 5.5.3.1.6.2.3

Rewrite the expression.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} + \frac{- 16 y^{2}}{x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.7

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} + \frac{- 80 y}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.8

Cancel the common factor of $- 80$ and $5$ .

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Step 5.5.3.1.8.1

Factor $5$ out of $- 80 y$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} + \frac{5 (- 16 y)}{5 x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.8.2

Cancel the common factors.

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Step 5.5.3.1.8.2.1

Factor $5$ out of $5 x {(y + 2)}^{4}$ .

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} + \frac{5 (- 16 y)}{5 (x {(y + 2)}^{4})} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.8.2.2

Cancel the common factor.

Step 5.5.3.1.8.2.3

Rewrite the expression.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} + \frac{- 16 y}{x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.9

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} - \frac{16 y}{x {(y + 2)}^{4}} + \frac{- 32}{5 x {(y + 2)}^{4}}$

Step 5.5.3.1.10

Move the negative in front of the fraction.

$y' = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} - \frac{16 y}{x {(y + 2)}^{4}} - \frac{32}{5 x {(y + 2)}^{4}}$

Step 6

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

$\frac{d y}{d x} = - \frac{y^{5}}{5 x {(y + 2)}^{4}} - \frac{2 y^{4}}{x {(y + 2)}^{4}} - \frac{8 y^{3}}{x {(y + 2)}^{4}} - \frac{16 y^{2}}{x {(y + 2)}^{4}} - \frac{16 y}{x {(y + 2)}^{4}} - \frac{32}{5 x {(y + 2)}^{4}}$