Calculus Examples

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{4}) = \frac{d}{d x} (3 x y - 5)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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Step 2.1.1

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

$\frac{d}{d u} [u^{4}] \frac{d}{d x} [y]$

Step 2.1.2

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

$4 u^{3} \frac{d}{d x} [y]$

Step 2.1.3

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

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

Step 2.2

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

$4 y^{3} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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 $g (x) = y$ .

$3 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [- 5]$

Step 3.2.3

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

$3 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [- 5]$

Step 3.2.4

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

$3 (x y' + y \cdot 1) + \frac{d}{d x} [- 5]$

Step 3.2.5

Multiply $y$ by $1$ .

$3 (x y' + y) + \frac{d}{d x} [- 5]$

Step 3.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$3 (x y' + y) + 0$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$3 (x y') + 3 y + 0$

Step 3.4.2

Add $3 x y' + 3 y$ and $0$ .

$3 x y' + 3 y$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $3 x y'$ from both sides of the equation.

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

Step 5.2

Factor $y'$ out of $4 y^{3} y' - 3 x y'$ .

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Step 5.2.1

Factor $y'$ out of $4 y^{3} y'$ .

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

Step 5.2.2

Factor $y'$ out of $- 3 x y'$ .

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4 y^{3} - 3 x$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 6

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

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