Calculus Examples

$4 x \cos (y) + 7 \sin (2 y) = 5 \sin (y)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (4 x \cos (y) + 7 \sin (2 y)) = \frac{d}{d x} (5 \sin (y))$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $4 x \cos (y) + 7 \sin (2 y)$ with respect to $x$ is $\frac{d}{d x} [4 x \cos (y)] + \frac{d}{d x} [7 \sin (2 y)]$ .

$\frac{d}{d x} [4 x \cos (y)] + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2

Evaluate $\frac{d}{d x} [4 x \cos (y)]$ .

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Step 2.2.1

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

$4 \frac{d}{d x} [x \cos (y)] + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \cos (y)$ .

$4 (x \frac{d}{d x} [\cos (y)] + \cos (y) \frac{d}{d x} [x]) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.3

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

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$4 (x (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d x} [y]) + \cos (y) \frac{d}{d x} [x]) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.3.2

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

$4 (x (- \sin (u_{1}) \frac{d}{d x} [y]) + \cos (y) \frac{d}{d x} [x]) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $y$ .

$4 (x (- \sin (y) \frac{d}{d x} [y]) + \cos (y) \frac{d}{d x} [x]) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.4

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

$4 (x (- \sin (y) y') + \cos (y) \frac{d}{d x} [x]) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.5

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

$4 (x (- \sin (y) y') + \cos (y) \cdot 1) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.2.6

Multiply $\cos (y)$ by $1$ .

$4 (x (- \sin (y) y') + \cos (y)) + \frac{d}{d x} [7 \sin (2 y)]$

Step 2.3

Evaluate $\frac{d}{d x} [7 \sin (2 y)]$ .

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Step 2.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7 \sin (2 y)$ with respect to $x$ is $7 \frac{d}{d x} [\sin (2 y)]$ .

$4 (x (- \sin (y) y') + \cos (y)) + 7 \frac{d}{d x} [\sin (2 y)]$

Step 2.3.2

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

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Step 2.3.2.1

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

$4 (x (- \sin (y) y') + \cos (y)) + 7 (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [2 y])$

Step 2.3.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$4 (x (- \sin (y) y') + \cos (y)) + 7 (\cos (u_{2}) \frac{d}{d x} [2 y])$

Step 2.3.2.3

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

$4 (x (- \sin (y) y') + \cos (y)) + 7 (\cos (2 y) \frac{d}{d x} [2 y])$

Step 2.3.3

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

$4 (x (- \sin (y) y') + \cos (y)) + 7 (\cos (2 y) (2 \frac{d}{d x} [y]))$

Step 2.3.4

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

$4 (x (- \sin (y) y') + \cos (y)) + 7 (\cos (2 y) (2 y'))$

Step 2.3.5

Multiply $2$ by $7$ .

$4 (x (- \sin (y) y') + \cos (y)) + 14 \cos (2 y) y'$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$4 (x (- \sin (y) y')) + 4 \cos (y) + 14 \cos (2 y) y'$

Step 2.4.2

Multiply $- 1$ by $4$ .

$- 4 x \sin (y) y' + 4 \cos (y) + 14 \cos (2 y) y'$

Step 2.4.3

Reorder terms.

$- 4 x \sin (y) y' + 14 \cos (2 y) y' + 4 \cos (y)$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

$5 \frac{d}{d x} [\sin (y)]$

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 $f (x) = \sin (x)$ and $g (x) = y$ .

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Step 3.2.1

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

$5 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [y])$

Step 3.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$5 (\cos (u) \frac{d}{d x} [y])$

Step 3.2.3

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

$5 (\cos (y) \frac{d}{d x} [y])$

Step 3.3

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

$5 \cos (y) y'$

Step 4

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

$- 4 x \sin (y) y' + 14 \cos (2 y) y' + 4 \cos (y) = 5 \cos (y) y'$

Step 5

Solve for $y'$ .

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Step 5.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$- 4 x \sin (y) y' + 14 (1 - 2 \sin^{2} (y)) y' + 4 \cos (y) = 5 \cos (y) y'$

Step 5.2

Subtract $4 \cos (y)$ from both sides of the equation.

$- 4 x \sin (y) y' + 14 (1 - 2 \sin^{2} (y)) y' = 5 \cos (y) y' - 4 \cos (y)$

Step 5.3

Subtract $5 \cos (y) y'$ from both sides of the equation.

$- 4 x \sin (y) y' + 14 (1 - 2 \sin^{2} (y)) y' - 5 \cos (y) y' = - 4 \cos (y)$

Step 5.4

Simplify the left side.

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Step 5.4.1

Simplify $- 4 x \sin (y) y' + 14 (1 - 2 \sin^{2} (y)) y' - 5 \cos (y) y'$ .

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Step 5.4.1.1

Simplify each term.

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Step 5.4.1.1.1

Apply the distributive property.

$- 4 x \sin (y) y' + (14 \cdot 1 + 14 (- 2 \sin^{2} (y))) y' - 5 \cos (y) y' = - 4 \cos (y)$

Step 5.4.1.1.2

Multiply $14$ by $1$ .

$- 4 x \sin (y) y' + (14 + 14 (- 2 \sin^{2} (y))) y' - 5 \cos (y) y' = - 4 \cos (y)$

Step 5.4.1.1.3

Multiply $- 2$ by $14$ .

$- 4 x \sin (y) y' + (14 - 28 \sin^{2} (y)) y' - 5 \cos (y) y' = - 4 \cos (y)$

Step 5.4.1.1.4

Apply the distributive property.

$- 4 x \sin (y) y' + 14 y' - 28 \sin^{2} (y) y' - 5 \cos (y) y' = - 4 \cos (y)$

Step 5.4.1.2

Reorder factors in $- 4 x \sin (y) y' + 14 y' - 28 \sin^{2} (y) y' - 5 \cos (y) y'$ .

$- 4 x y' \sin (y) + 14 y' - 28 y' \sin^{2} (y) - 5 y' \cos (y) = - 4 \cos (y)$

Step 5.5

Solve the equation for $y'$ .

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Step 5.5.1

Factor $y'$ out of $- 4 x y' \sin (y) + 14 y' - 28 y' \sin^{2} (y) - 5 y' \cos (y)$ .

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Step 5.5.1.1

Factor $y'$ out of $- 4 x y' \sin (y)$ .

$y' (- 4 x \sin (y)) + 14 y' - 28 y' \sin^{2} (y) - 5 y' \cos (y) = - 4 \cos (y)$

Step 5.5.1.2

Factor $y'$ out of $14 y'$ .

$y' (- 4 x \sin (y)) + y' \cdot 14 - 28 y' \sin^{2} (y) - 5 y' \cos (y) = - 4 \cos (y)$

Step 5.5.1.3

Factor $y'$ out of $- 28 y' \sin^{2} (y)$ .

$y' (- 4 x \sin (y)) + y' \cdot 14 + y' (- 28 \sin^{2} (y)) - 5 y' \cos (y) = - 4 \cos (y)$

Step 5.5.1.4

Factor $y'$ out of $- 5 y' \cos (y)$ .

$y' (- 4 x \sin (y)) + y' \cdot 14 + y' (- 28 \sin^{2} (y)) + y' (- 5 \cos (y)) = - 4 \cos (y)$

Step 5.5.1.5

Factor $y'$ out of $y' (- 4 x \sin (y)) + y' \cdot 14$ .

$y' (- 4 x \sin (y) + 14) + y' (- 28 \sin^{2} (y)) + y' (- 5 \cos (y)) = - 4 \cos (y)$

Step 5.5.1.6

Factor $y'$ out of $y' (- 4 x \sin (y) + 14) + y' (- 28 \sin^{2} (y))$ .

$y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y)) + y' (- 5 \cos (y)) = - 4 \cos (y)$

Step 5.5.1.7

Factor $y'$ out of $y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y)) + y' (- 5 \cos (y))$ .

$y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)) = - 4 \cos (y)$

Step 5.5.2

Divide each term in $y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)) = - 4 \cos (y)$ by $- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)$ and simplify.

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Step 5.5.2.1

Divide each term in $y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)) = - 4 \cos (y)$ by $- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)$ .

$\frac{y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y))}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)} = \frac{- 4 \cos (y)}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Cancel the common factor of $- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)$ .

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Step 5.5.2.2.1.1

Cancel the common factor.

$\frac{y' (- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y))}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)} = \frac{- 4 \cos (y)}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 4 \cos (y)}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

Move the negative in front of the fraction.

$y' = - \frac{4 \cos (y)}{- 4 x \sin (y) + 14 - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.3.2

Factor $- 1$ out of $- 4 x \sin (y)$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y)) + 14 - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.3.3

Rewrite $14$ as $- 1 (- 14)$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y)) - 1 (- 14) - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.3.4

Factor $- 1$ out of $- (4 x \sin (y)) - 1 (- 14)$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y) - 14) - 28 \sin^{2} (y) - 5 \cos (y)}$

Step 5.5.2.3.5

Factor $- 1$ out of $- 28 \sin^{2} (y)$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y) - 14) - (28 \sin^{2} (y)) - 5 \cos (y)}$

Step 5.5.2.3.6

Factor $- 1$ out of $- (4 x \sin (y) - 14) - (28 \sin^{2} (y))$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y) - 14 + 28 \sin^{2} (y)) - 5 \cos (y)}$

Step 5.5.2.3.7

Factor $- 1$ out of $- 5 \cos (y)$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y) - 14 + 28 \sin^{2} (y)) - (5 \cos (y))}$

Step 5.5.2.3.8

Factor $- 1$ out of $- (4 x \sin (y) - 14 + 28 \sin^{2} (y)) - (5 \cos (y))$ .

$y' = - \frac{4 \cos (y)}{- (4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y))}$

Step 5.5.2.3.9

Simplify the expression.

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Step 5.5.2.3.9.1

Rewrite $- (4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y))$ as $- 1 (4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y))$ .

$y' = - \frac{4 \cos (y)}{- 1 (4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y))}$

Step 5.5.2.3.9.2

Move the negative in front of the fraction.

$y' = - - \frac{4 \cos (y)}{4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y)}$

Step 5.5.2.3.9.3

Multiply $- 1$ by $- 1$ .

$y' = 1 \frac{4 \cos (y)}{4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y)}$

Step 5.5.2.3.9.4

Multiply $\frac{4 \cos (y)}{4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y)}$ by $1$ .

$y' = \frac{4 \cos (y)}{4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y)}$

Step 6

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

$\frac{d y}{d x} = \frac{4 \cos (y)}{4 x \sin (y) - 14 + 28 \sin^{2} (y) + 5 \cos (y)}$