Calculus Examples

$\tan (x + y) = x$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\tan (x + y)) = \frac{d}{d x} (x)$

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 $f (x) = \tan (x)$ and $g (x) = x + y$ .

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Step 2.1.1

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

$\frac{d}{d u} [\tan (u)] \frac{d}{d x} [x + y]$

Step 2.1.2

The derivative of $\tan (u)$ with respect to $u$ is $\sec^{2} (u)$ .

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

$\sec^{2} (x + y) (1 + y')$

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.

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

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

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Step 5.1.2.1.1

Cancel the common factor.

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

Step 5.1.2.1.2

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

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

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Rewrite $1$ as $1^{2}$ .

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

Step 5.1.3.2

Rewrite $\frac{1^{2}}{\sec^{2} (x + y)}$ as ${(\frac{1}{\sec (x + y)})}^{2}$ .

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

Step 5.1.3.3

Rewrite $\sec (x + y)$ in terms of sines and cosines.

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

Step 5.1.3.4

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\cos (x + y)}$ .

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

Step 5.1.3.5

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

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

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 $1$ from both sides of the equation.

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

Step 5.2.2

Reorder $\cos^{2} (x + y)$ and $- 1$ .

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

Step 5.2.3

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

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

Step 5.2.4

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

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

Step 5.2.5

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

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

Step 5.2.6

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

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

Step 5.2.7

Apply pythagorean identity.

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

Step 6

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

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