Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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 $y^{3} + y^{2} - 5 y - x^{2}$ with respect to $y$ is $\frac{d}{d y} [y^{3}] + \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 5 y] + \frac{d}{d y} [- x^{2}]$ .

$\frac{d}{d y} [y^{3}] + \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 5 y] + \frac{d}{d y} [- x^{2}]$

Step 2.1.2

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

$3 y^{2} + \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 5 y] + \frac{d}{d y} [- x^{2}]$

Step 2.1.3

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

$3 y^{2} + 2 y + \frac{d}{d y} [- 5 y] + \frac{d}{d y} [- x^{2}]$

Step 2.2

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

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Step 2.2.1

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

$3 y^{2} + 2 y - 5 \frac{d}{d y} [y] + \frac{d}{d y} [- x^{2}]$

Step 2.2.2

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

$3 y^{2} + 2 y - 5 \cdot 1 + \frac{d}{d y} [- x^{2}]$

Step 2.2.3

Multiply $- 5$ by $1$ .

$3 y^{2} + 2 y - 5 + \frac{d}{d y} [- x^{2}]$

Step 2.3

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

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Step 2.3.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- x^{2}$ with respect to $y$ is $- \frac{d}{d y} [x^{2}]$ .

$3 y^{2} + 2 y - 5 - \frac{d}{d y} [x^{2}]$

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $- 1$ .

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

Step 2.4

Reorder terms.

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

Step 3

Since $- 4$ is constant with respect to $y$ , the derivative of $- 4$ with respect to $y$ is $0$ .

$0$

Step 4

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

$3 y^{2} - 2 x x' + 2 y - 5 = 0$

Step 5

Solve for $x'$ .

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Step 5.1

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

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Step 5.1.1

Subtract $3 y^{2}$ from both sides of the equation.

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

Step 5.1.2

Subtract $2 y$ from both sides of the equation.

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

Step 5.1.3

Add $5$ to both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $- 2$ .

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Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Rewrite the expression.

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

Step 5.2.2.2

Cancel the common factor of $x$ .

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Step 5.2.2.2.1

Cancel the common factor.

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

Step 5.2.2.2.2

Divide $x'$ by $1$ .

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

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Dividing two negative values results in a positive value.

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

Step 5.2.3.1.2

Cancel the common factor of $- 2$ .

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Step 5.2.3.1.2.1

Cancel the common factor.

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

Step 5.2.3.1.2.2

Rewrite the expression.

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

Step 5.2.3.1.3

Move the negative in front of the fraction.

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

Step 6

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

$\frac{d x}{d y} = \frac{3 y^{2}}{2 x} + \frac{y}{x} - \frac{5}{2 x}$