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)].