If y=cos(x+y), then dy/dx= -sin(x+y)/[1+sin(x+y)]. Here, we learn how to differentiate y=cos(x+y) with respect to x. Let us find the derivative of y=cos(x+y).

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

**Question:** If y=cos(x+y), then $\dfrac{dy}{dx}$.

**Solution:**

We are given that

y = cos(x+y).

**Step 1:** Differentiating both sides of the above equation with respect x, we get that

$\dfrac{dy}{dx}$ = $-\sin(x+y)\dfrac{d}{dx}(x+y)$, by the chain rule.

⇒ $\dfrac{dy}{dx}$ = $-\sin(x+y)(1+\dfrac{dy}{dx})$

**Step 2:** We now solve the above equation for dy/dx.

$\dfrac{dy}{dx}$ = $-\sin(x+y)-\sin(x+y)\dfrac{dy}{dx}$

⇒ $\dfrac{dy}{dx} + \sin(x+y)\dfrac{dy}{dx}$ = $-\sin(x+y)$

⇒ $[1+\sin(x+y)]\dfrac{dy}{dx}$ = $-\sin(x+y)$

⇒ $\dfrac{dy}{dx} = -\dfrac{\sin(x+y)}{1+\sin(x+y)}$.

Therefore, if y=cos(x+y), then dy/dx= -sin(x+y)/[1+sin(x+y)].

**Related Questions:**

Derivative of $\dfrac{1}{\sin x}$

Derivative of $\dfrac{\sin x}{x}$

## FAQs

### Q1: If y=cos(x+y), then find dy/dx.

Answer: If y=cos(x+y), then dy/dx= -sin(x+y)/[1+sin(x+y)].