Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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 $g (x) = x^{2} + y$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

Replace all occurrences of $u$ with $x^{2} + y$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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

Step 3

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

$1$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Dividing two negative values results in a positive value.

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

Step 5.1.2.2

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

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Step 5.1.2.2.1

Cancel the common factor.

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

Step 5.1.2.2.2

Divide $2 x + y'$ by $1$ .

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

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Cancel the common factor of $1$ and $- 1$ .

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Step 5.1.3.1.1

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

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

Step 5.1.3.1.2

Move the negative in front of the fraction.

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

Step 5.1.3.2

Convert from $\frac{1}{\sin (x^{2} + y)}$ to $\csc (x^{2} + y)$ .

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

Step 5.2

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

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

Step 6

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

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