Find dy/dx y=cos(x+y)

$y = \cos (x + y)$

Differentiate both sides of the equation.

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

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

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 $g (x) = x + y$ .

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

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

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

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

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

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

Differentiate.

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

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

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

$- \sin (x + y) (1 + \frac{d}{d x} [y])$

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

$- \sin (x + y) (1 + \frac{d}{d x} [y])$

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

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

Solve for $y'$ .

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Simplify $- \sin (x + y) (1 + y')$ .

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Apply the distributive property.

$y' = - \sin (x + y) \cdot 1 - \sin (x + y) y'$

Simplify the expression.

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Multiply $- 1$ by $1$ .

$y' = - \sin (x + y) - \sin (x + y) y'$

Reorder factors in $- \sin (x + y) - \sin (x + y) y'$ .

$y' = - \sin (x + y) - y' (\sin (x + y))$

Add $- y' (\sin (x + y))$ to both sides of the equation.

$y' + y' (\sin (x + y)) = - \sin (x + y)$

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

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Factor $y'$ out of ${y'}^{1}$ .

$y' \cdot 1 + y' (\sin (x + y)) = - \sin (x + y)$

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

$y' \cdot 1 + y' (\sin (x + y)) = - \sin (x + y)$

Factor $y'$ out of $y' \cdot 1 + y' \sin (x + y)$ .

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

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

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

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

Cancel the common factor of $1 + \sin (x + y)$ .

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Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Move the negative in front of the fraction.

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

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

$\frac{d y}{d x} = - \frac{\sin (x + y)}{1 + \sin (x + y)}$

Find dy/dx y=cos(x+y)

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