Question

How do you find the derivative of `sin (cos(2t-5))`?

Answer 1

`(df)/(dt)=-2sin(2t-5)cos(cos(2t-5))`

Explanation:

To find the derivative of `sin(cos(2t−5))`, we use the concept of function of a function and use chain rule to find derivative.

Here `f(t)=sint`, `g(t)=cost` and `h(t)=(2t-5)`

Hence `f(g(h(t)))=sin(cos(2t−5))`

as `(df)/(dt)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dt)`

Hence `(df)/(dt)=cos(cos(2t-5))xx(-sin(2t-5))xx2`

or `(df)/(dt)=-2sin(2t-5)cos(cos(2t-5))`