Calculus Examples

$y^{3} + 4 x^{2} + 1 = - 5 x - 3 y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{3} + 4 x^{2} + 1) = \frac{d}{d x} (- 5 x - 3 y)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $y^{3} + 4 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [y^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [y^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 2.2

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

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Step 2.2.1

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

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Step 2.2.1.1

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

$\frac{d}{d u} [u^{3}] \frac{d}{d x} [y] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 2.2.1.2

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

$3 u^{2} \frac{d}{d x} [y] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 2.2.1.3

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

$3 y^{2} \frac{d}{d x} [y] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 2.2.2

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

$3 y^{2} y' + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [1]$

Step 2.3

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Step 2.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$3 y^{2} y' + 4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 2.3.2

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

$3 y^{2} y' + 4 (2 x) + \frac{d}{d x} [1]$

Step 2.3.3

Multiply $2$ by $4$ .

$3 y^{2} y' + 8 x + \frac{d}{d x} [1]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$3 y^{2} y' + 8 x + 0$

Step 2.4.2

Add $3 y^{2} y' + 8 x$ and $0$ .

$3 y^{2} y' + 8 x$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $- 5$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$- 5 - 3 y'$

Step 4

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

$3 y^{2} y' + 8 x = - 5 - 3 y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $3 y'$ to both sides of the equation.

$3 y^{2} y' + 8 x + 3 y' = - 5$

Step 5.2

Subtract $8 x$ from both sides of the equation.

$3 y^{2} y' + 3 y' = - 5 - 8 x$

Step 5.3

Factor $3 y'$ out of $3 y^{2} y' + 3 y'$ .

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Step 5.3.1

Factor $3 y'$ out of $3 y^{2} y'$ .

$3 y' y^{2} + 3 y' = - 5 - 8 x$

Step 5.3.2

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

$3 y' y^{2} + 3 y' \cdot 1 = - 5 - 8 x$

Step 5.3.3

Factor $3 y'$ out of $3 y' y^{2} + 3 y' \cdot 1$ .

$3 y' (y^{2} + 1) = - 5 - 8 x$

Step 5.4

Divide each term in $3 y' (y^{2} + 1) = - 5 - 8 x$ by $3 (y^{2} + 1)$ and simplify.

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Step 5.4.1

Divide each term in $3 y' (y^{2} + 1) = - 5 - 8 x$ by $3 (y^{2} + 1)$ .

$\frac{3 y' (y^{2} + 1)}{3 (y^{2} + 1)} = \frac{- 5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{3 y' (y^{2} + 1)}{3 (y^{2} + 1)} = \frac{- 5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (y^{2} + 1)}{y^{2} + 1} = \frac{- 5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.2.2

Cancel the common factor of $y^{2} + 1$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (y^{2} + 1)}{y^{2} + 1} = \frac{- 5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{5}{3 (y^{2} + 1)} + \frac{- 8 x}{3 (y^{2} + 1)}$

Step 5.4.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{5}{3 (y^{2} + 1)} - \frac{8 x}{3 (y^{2} + 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{5}{3 (y^{2} + 1)} - \frac{8 x}{3 (y^{2} + 1)}$