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

$x = \cos (y)$

Differentiate both sides of the equation.

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

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

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

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

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

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

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

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

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

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

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

$1 = - \sin (y) y'$

Solve for $y'$ .

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Rewrite the equation as $- \sin (y) y' = 1$ .

$- \sin (y) y' = 1$

Divide each term by $- 1 \sin (y)$ and simplify.

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Divide each term in $- \sin (y) y' = 1$ by $- 1 \sin (y)$ .

$\frac{- \sin (y) y'}{- 1 \sin (y)} = \frac{1}{- 1 \sin (y)}$

Simplify the left side of the equation by multiplying out all the terms.

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Dividing two negative values results in a positive value.

$\frac{\sin (y) y'}{\sin (y)} = \frac{1}{- 1 \sin (y)}$

Cancel the common factor of $\sin (y)$ .

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Cancel the common factor.

$\frac{\sin (y) y'}{\sin (y)} = \frac{1}{- 1 \sin (y)}$

Divide $y'$ by $1$ .

$y' = \frac{1}{- 1 \sin (y)}$

Simplify $\frac{1}{- 1 \sin (y)}$ .

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

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

$y' = \frac{- 1 \cdot - 1}{- 1 \sin (y)}$

Move the negative in front of the fraction.

$y' = - \frac{1}{\sin (y)}$

Convert from $\frac{1}{\sin (y)}$ to $\csc (y)$ .

$y' = - \csc (y)$

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

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

Find dy/dx x=cos(y)

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