Question

derivative of (cos5x+cos7x)​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{d}{dx}\:[\:\cos\:(\:5x\:)\:+\:\cos\:(\:7x\:)\:=\:-\:5\:\sin\:(\:5x\:)\:-\:7\:\sin\:(\:7x\:)\:}}}$

Step-by-step-explanation:

We have given a trigonometric function.

We have to find the derivative of that function.

The given trigonometric function is

$\displaystyle{\sf\:\cos\:(\:5x\:)\:+\:\cos\:(\:7x\:)}$

Let this function be y.

$\displaystyle{\therefore\:y\:=\:\cos\:(\:5x\:)\:+\:\cos\:(\:7x\:)}$

Differentiating both sides w.r.t. x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:y\:)\:=\:\dfrac{d}{dx}\:\left[\:\cos\:(\:5x\:)\:+\:\cos\:(\:7x\:)\:\right]}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:(\:u\:+\:v\:)\:=\:\dfrac{du}{dx}\:+\:\dfrac{dv}{dx}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:[\:\cos\:(\:5x\:)\:]\:+\:\dfrac{d}{dx}\:[\:\cos\:(\:7x\:)\:]}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d}{dx}\:[\:\cos\:(\:x\:)\:]\:=\:-\:\sin\:(\:x\:)\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:-\:\sin\:(\:5x\:)\:.\:\dfrac{d}{dx}\:(\:5x\:)\:+\:[\:-\:\sin\:(\:7x\:)\:]\:.\:\dfrac{d}{dx}\:(\:7x\:)}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\dfrac{d}{dx}\:(\:kx\:)\:=\:k\:\dfrac{d}{dx}\:(\:x\:)\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:-\:\sin\:(\:5x\:)\:5\:\dfrac{d}{dx}\:(\:x\:)\:-\:\sin\:(\:7x\:)\:7\:\dfrac{d}{dx}\:(\:x\:)}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:-\:5\:\sin\:(\:5x\:)\:\times\:1\:-\:7\:\sin\:(\:7x\:)\:\times\:1}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:-\:5\:\sin\:(\:5x\:)\:-\:7\:\sin\:(\:7x\:)}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{d}{dx}\:[\:\cos\:(\:5x\:)\:+\:\cos\:(\:7x\:)\:=\:-\:5\:\sin\:(\:5x\:)\:-\:7\:\sin\:(\:7x\:)\:}}}}$