Calculus Examples

$x^{5} + 32 y^{3} = \ln (y)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{5} + 32 y^{3}) = \frac{d}{d x} (\ln (y))$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

$\frac{d}{d x} [x^{5}] + \frac{d}{d x} [32 y^{3}]$

Step 2.1.2

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

$5 x^{4} + \frac{d}{d x} [32 y^{3}]$

Step 2.2

Evaluate $\frac{d}{d x} [32 y^{3}]$ .

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Step 2.2.1

Since $32$ is constant with respect to $x$ , the derivative of $32 y^{3}$ with respect to $x$ is $32 \frac{d}{d x} [y^{3}]$ .

$5 x^{4} + 32 \frac{d}{d x} [y^{3}]$

Step 2.2.2

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

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $y$ .

$5 x^{4} + 32 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 2.2.2.2

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

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

Step 2.2.2.3

Replace all occurrences of $u$ with $y$ .

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

Step 2.2.3

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

$5 x^{4} + 32 (3 y^{2} y')$

Step 2.2.4

Multiply $3$ by $32$ .

$5 x^{4} + 96 y^{2} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 $g (x) = y$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [y]$

Step 3.1.2

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

$\frac{1}{u} \frac{d}{d x} [y]$

Step 3.1.3

Replace all occurrences of $u$ with $y$ .

$\frac{1}{y} \frac{d}{d x} [y]$

Step 3.2

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

$\frac{1}{y} y'$

Step 3.3

Combine $\frac{1}{y}$ and $y'$ .

$\frac{y'}{y}$

Step 4

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

$5 x^{4} + 96 y^{2} y' = \frac{y'}{y}$

Step 5

Solve for $y'$ .

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Step 5.1

Multiply both sides by $y$ .

$(5 x^{4} + 96 y^{2} y') y = \frac{y'}{y} y$

Step 5.2

Simplify.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Simplify $(5 x^{4} + 96 y^{2} y') y$ .

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Step 5.2.1.1.1

Apply the distributive property.

$5 x^{4} y + 96 y^{2} y' y = \frac{y'}{y} y$

Step 5.2.1.1.2

Multiply $y^{2}$ by $y$ by adding the exponents.

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Step 5.2.1.1.2.1

Move $y$ .

$5 x^{4} y + 96 (y \cdot y^{2}) y' = \frac{y'}{y} y$

Step 5.2.1.1.2.2

Multiply $y$ by $y^{2}$ .

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Step 5.2.1.1.2.2.1

Raise $y$ to the power of $1$ .

$5 x^{4} y + 96 (y^{1} y^{2}) y' = \frac{y'}{y} y$

Step 5.2.1.1.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$5 x^{4} y + 96 y^{1 + 2} y' = \frac{y'}{y} y$

Step 5.2.1.1.2.3

Add $1$ and $2$ .

$5 x^{4} y + 96 y^{3} y' = \frac{y'}{y} y$

Step 5.2.1.1.3

Move $y^{3}$ .

$5 x^{4} y + 96 y' y^{3} = \frac{y'}{y} y$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Cancel the common factor of $y$ .

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Step 5.2.2.1.1

Cancel the common factor.

$5 x^{4} y + 96 y' y^{3} = \frac{y'}{y} y$

Step 5.2.2.1.2

Rewrite the expression.

$5 x^{4} y + 96 y' y^{3} = y'$

Step 5.3

Solve for $y'$ .

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Step 5.3.1

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

$5 x^{4} y + 96 y' y^{3} - y' = 0$

Step 5.3.2

Subtract $5 x^{4} y$ from both sides of the equation.

$96 y' y^{3} - y' = - 5 x^{4} y$

Step 5.3.3

Factor $y'$ out of $96 y' y^{3} - y'$ .

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Step 5.3.3.1

Factor $y'$ out of $96 y' y^{3}$ .

$y' (96 y^{3}) - y' = - 5 x^{4} y$

Step 5.3.3.2

Factor $y'$ out of $- y'$ .

$y' (96 y^{3}) + y' \cdot - 1 = - 5 x^{4} y$

Step 5.3.3.3

Factor $y'$ out of $y' (96 y^{3}) + y' \cdot - 1$ .

$y' (96 y^{3} - 1) = - 5 x^{4} y$

Step 5.3.4

Divide each term in $y' (96 y^{3} - 1) = - 5 x^{4} y$ by $96 y^{3} - 1$ and simplify.

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Step 5.3.4.1

Divide each term in $y' (96 y^{3} - 1) = - 5 x^{4} y$ by $96 y^{3} - 1$ .

$\frac{y' (96 y^{3} - 1)}{96 y^{3} - 1} = \frac{- 5 x^{4} y}{96 y^{3} - 1}$

Step 5.3.4.2

Simplify the left side.

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Step 5.3.4.2.1

Cancel the common factor of $96 y^{3} - 1$ .

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Step 5.3.4.2.1.1

Cancel the common factor.

$\frac{y' (96 y^{3} - 1)}{96 y^{3} - 1} = \frac{- 5 x^{4} y}{96 y^{3} - 1}$

Step 5.3.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 5 x^{4} y}{96 y^{3} - 1}$

Step 5.3.4.3

Simplify the right side.

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Step 5.3.4.3.1

Move the negative in front of the fraction.

$y' = - \frac{5 x^{4} y}{96 y^{3} - 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{5 x^{4} y}{96 y^{3} - 1}$