Find dy/dx x=tan(y) - MathMaster

$x = \tan (y)$

Differentiate both sides of the equation.

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

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

$1$

Differentiate the right side of the equation.

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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 $g (x) = y$ .

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

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

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

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

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

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

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

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

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

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

Solve for $y'$ .

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Rewrite the equation as $\sec^{2} (y) y' = 1$ .

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

Divide each term by $\sec^{2} (y)$ and simplify.

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

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

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

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Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify $\frac{1}{\sec^{2} (y)}$ .

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

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

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

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

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

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

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

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

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

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

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

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

Find dy/dx x=tan(y)

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