Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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

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

Step 2.1.2

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

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

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 $y$ .

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

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 $n = 3$ .

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $- 1$ by $1$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.5

Reorder terms.

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

Step 3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$0$

Step 4

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

$3 y^{2} y' + 2 x + 5 y' - 1 = 0$

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 $2 x$ from both sides of the equation.

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

Step 5.1.2

Add $1$ to both sides of the equation.

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

Step 5.2

Factor $y'$ out of $3 y^{2} y' + 5 y'$ .

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y'$ out of $y' (3 y^{2}) + y' \cdot 5$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $3 y^{2} + 5$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2

Factor $- 1$ out of $- 2 x$ .

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

Step 5.3.3.3

Rewrite $1$ as $- 1 (- 1)$ .

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

Step 5.3.3.4

Factor $- 1$ out of $- (2 x) - 1 (- 1)$ .

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

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (2 x - 1)$ as $- 1 (2 x - 1)$ .

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

Step 5.3.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

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