Calculus Examples

$\cos (2 x + 3 y^{3}) = 3 y^{2}$

Step 1

Differentiate both sides of the equation.

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

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 $f (x) = \cos (x)$ and $g (x) = 2 x + 3 y^{3}$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $2 x + 3 y^{3}$ .

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

Step 2.1.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$- \sin (u_{1}) \frac{d}{d x} [2 x + 3 y^{3}]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $2 x + 3 y^{3}$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $1$ .

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

Step 2.2.5

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

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

Step 2.3

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 2.3.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

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

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

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

Step 2.3.3

Replace all occurrences of $u_{2}$ with $y$ .

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

Step 2.4

Multiply $3$ by $3$ .

$- \sin (2 x + 3 y^{3}) (2 + 9 (y^{2} \frac{d}{d x} [y]))$

Step 2.5

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

$- \sin (2 x + 3 y^{3}) (2 + 9 y^{2} y')$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $3 \frac{d}{d x} [y^{2}]$ .

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

Step 3.2

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 3.2.1

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

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

Step 3.2.2

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

$3 (2 u \frac{d}{d x} [y])$

Step 3.2.3

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

$3 (2 y \frac{d}{d x} [y])$

Step 3.3

Multiply $2$ by $3$ .

$6 (y \frac{d}{d x} [y])$

Step 3.4

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

$6 y y'$

Step 4

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

$- \sin (2 x + 3 y^{3}) (2 + 9 y^{2} y') = 6 y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $- \sin (2 x + 3 y^{3}) (2 + 9 y^{2} y')$ .

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Step 5.1.1.1

Rewrite.

$0 + 0 - \sin (2 x + 3 y^{3}) (2 + 9 y^{2} y') = 6 y y'$

Step 5.1.1.2

Simplify by adding zeros.

$- \sin (2 x + 3 y^{3}) (2 + 9 y^{2} y') = 6 y y'$

Step 5.1.1.3

Apply the distributive property.

$- \sin (2 x + 3 y^{3}) \cdot 2 - \sin (2 x + 3 y^{3}) (9 y^{2} y') = 6 y y'$

Step 5.1.1.4

Simplify the expression.

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Step 5.1.1.4.1

Multiply $2$ by $- 1$ .

$- 2 \sin (2 x + 3 y^{3}) - \sin (2 x + 3 y^{3}) (9 y^{2} y') = 6 y y'$

Step 5.1.1.4.2

Multiply $9$ by $- 1$ .

$- 2 \sin (2 x + 3 y^{3}) - 9 \sin (2 x + 3 y^{3}) (y^{2} y') = 6 y y'$

Step 5.1.1.4.3

Reorder factors in $- 2 \sin (2 x + 3 y^{3}) - 9 \sin (2 x + 3 y^{3}) (y^{2} y')$ .

$- 2 \sin (2 x + 3 y^{3}) - 9 y^{2} y' \sin (2 x + 3 y^{3}) = 6 y y'$

Step 5.2

Subtract $6 y y'$ from both sides of the equation.

$- 2 \sin (2 x + 3 y^{3}) - 9 y^{2} y' \sin (2 x + 3 y^{3}) - 6 y y' = 0$

Step 5.3

Add $2 \sin (2 x + 3 y^{3})$ to both sides of the equation.

$- 9 y^{2} y' \sin (2 x + 3 y^{3}) - 6 y y' = 2 \sin (2 x + 3 y^{3})$

Step 5.4

Factor $3 y y'$ out of $- 9 y^{2} y' \sin (2 x + 3 y^{3}) - 6 y y'$ .

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Step 5.4.1

Factor $3 y y'$ out of $- 9 y^{2} y' \sin (2 x + 3 y^{3})$ .

$3 y y' (- 3 y \sin (2 x + 3 y^{3})) - 6 y y' = 2 \sin (2 x + 3 y^{3})$

Step 5.4.2

Factor $3 y y'$ out of $- 6 y y'$ .

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

Step 5.4.3

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

$3 y y' (- 3 y \sin (2 x + 3 y^{3}) - 2) = 2 \sin (2 x + 3 y^{3})$

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $3$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Rewrite the expression.

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

Step 5.5.2.2

Cancel the common factor of $y$ .

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Step 5.5.2.2.1

Cancel the common factor.

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

Step 5.5.2.2.2

Rewrite the expression.

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

Step 5.5.2.3

Cancel the common factor of $- 3 y \sin (2 x + 3 y^{3}) - 2$ .

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Step 5.5.2.3.1

Cancel the common factor.

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

Step 5.5.2.3.2

Divide $y'$ by $1$ .

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

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Factor $- 1$ out of $- 3 y \sin (2 x + 3 y^{3})$ .

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

Step 5.5.3.2

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

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

Step 5.5.3.3

Factor $- 1$ out of $- (3 y \sin (2 x + 3 y^{3})) - 1 (2)$ .

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

Step 5.5.3.4

Rewrite negatives.

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Step 5.5.3.4.1

Rewrite $- (3 y \sin (2 x + 3 y^{3}) + 2)$ as $- 1 (3 y \sin (2 x + 3 y^{3}) + 2)$ .

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

Step 5.5.3.4.2

Move the negative in front of the fraction.

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

Step 5.5.3.4.3

Reorder factors in $- \frac{2 \sin (2 x + 3 y^{3})}{(3 y) (3 y \sin (2 x + 3 y^{3}) + 2)}$ .

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

Step 6

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

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