Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $5 x^{3} - x^{2} + 3 y$ with respect to $x$ is $\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [3 y]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [5 x^{3}]$ .

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Step 2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $5$ .

$15 x^{2} + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [3 y]$

Step 2.3

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

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Step 2.3.1

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

$15 x^{2} - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 y]$

Step 2.3.2

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

$15 x^{2} - (2 x) + \frac{d}{d x} [3 y]$

Step 2.3.3

Multiply $2$ by $- 1$ .

$15 x^{2} - 2 x + \frac{d}{d x} [3 y]$

Step 2.4

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

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Step 2.4.1

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

$15 x^{2} - 2 x + 3 \frac{d}{d x} [y]$

Step 2.4.2

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

$15 x^{2} - 2 x + 3 y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

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

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

Step 3.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [4 y^{2}]$

Step 3.2

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

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Step 3.2.1

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

$0 + 4 \frac{d}{d x} [y^{2}]$

Step 3.2.2

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

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Step 3.2.2.1

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

$0 + 4 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

Step 3.2.2.2

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

$0 + 4 (2 u \frac{d}{d x} [y])$

Step 3.2.2.3

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

$0 + 4 (2 y \frac{d}{d x} [y])$

Step 3.2.3

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

$0 + 4 (2 y y')$

Step 3.2.4

Multiply $2$ by $4$ .

$0 + 8 y y'$

Step 3.3

Add $0$ and $8 y y'$ .

$8 y y'$

Step 4

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

$15 x^{2} - 2 x + 3 y' = 8 y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $8 y y'$ from both sides of the equation.

$15 x^{2} - 2 x + 3 y' - 8 y y' = 0$

Step 5.2

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

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Step 5.2.1

Subtract $15 x^{2}$ from both sides of the equation.

$- 2 x + 3 y' - 8 y y' = - 15 x^{2}$

Step 5.2.2

Add $2 x$ to both sides of the equation.

$3 y' - 8 y y' = - 15 x^{2} + 2 x$

Step 5.3

Factor $y'$ out of $3 y' - 8 y y'$ .

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Step 5.3.1

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

$y' \cdot 3 - 8 y y' = - 15 x^{2} + 2 x$

Step 5.3.2

Factor $y'$ out of $- 8 y y'$ .

$y' \cdot 3 + y' (- 8 y) = - 15 x^{2} + 2 x$

Step 5.3.3

Factor $y'$ out of $y' \cdot 3 + y' (- 8 y)$ .

$y' (3 - 8 y) = - 15 x^{2} + 2 x$

Step 5.4

Divide each term in $y' (3 - 8 y) = - 15 x^{2} + 2 x$ by $3 - 8 y$ and simplify.

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Step 5.4.1

Divide each term in $y' (3 - 8 y) = - 15 x^{2} + 2 x$ by $3 - 8 y$ .

$\frac{y' (3 - 8 y)}{3 - 8 y} = \frac{- 15 x^{2}}{3 - 8 y} + \frac{2 x}{3 - 8 y}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3 - 8 y$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (3 - 8 y)}{3 - 8 y} = \frac{- 15 x^{2}}{3 - 8 y} + \frac{2 x}{3 - 8 y}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 15 x^{2}}{3 - 8 y} + \frac{2 x}{3 - 8 y}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 15 x^{2} + 2 x}{3 - 8 y}$

Step 5.4.3.2

Factor $x$ out of $- 15 x^{2} + 2 x$ .

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Step 5.4.3.2.1

Factor $x$ out of $- 15 x^{2}$ .

$y' = \frac{x (- 15 x) + 2 x}{3 - 8 y}$

Step 5.4.3.2.2

Factor $x$ out of $2 x$ .

$y' = \frac{x (- 15 x) + x \cdot 2}{3 - 8 y}$

Step 5.4.3.2.3

Factor $x$ out of $x (- 15 x) + x \cdot 2$ .

$y' = \frac{x (- 15 x + 2)}{3 - 8 y}$

Step 5.4.3.3

Factor $- 1$ out of $- 15 x$ .

$y' = \frac{x (- (15 x) + 2)}{3 - 8 y}$

Step 5.4.3.4

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

$y' = \frac{x (- (15 x) - 1 (- 2))}{3 - 8 y}$

Step 5.4.3.5

Factor $- 1$ out of $- (15 x) - 1 (- 2)$ .

$y' = \frac{x (- (15 x - 2))}{3 - 8 y}$

Step 5.4.3.6

Simplify the expression.

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Step 5.4.3.6.1

Rewrite $- (15 x - 2)$ as $- 1 (15 x - 2)$ .

$y' = \frac{x (- 1 (15 x - 2))}{3 - 8 y}$

Step 5.4.3.6.2

Move the negative in front of the fraction.

$y' = - \frac{x (15 x - 2)}{3 - 8 y}$

Step 6

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

$\frac{d y}{d x} = - \frac{x (15 x - 2)}{3 - 8 y}$