Question

How do you differentiate `y =-(e^(x-sin^2x)` using the chain rule?

Answer 1

`(dy)/(dx)=-(1-sin2x)e^(x-sin^2x)`

Explanation:

In order to differentiate a function of a function, say `y, =f(g(x))`, where we have to find `(dy)/(dx)`, we need to do (a) substitute `u=g(x)`, which gives us `y=f(u)`. Then we need to use a formula called Chain Rule, which states that `(dy)/(dx)=(dy)/(du)xx(du)/(dx)`. In fact if we have something like `y=f(g(h(x)))`, we can have `(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)`

Hence for `y=-e^(x-sin^2x)`

`(dy)/(dx)=-e^(x-sin^2x)xxd/(dx)(x-sin^2x)`

= `-e^(x-sin^2x)xx(1-2sinx xxd/(dx)sinx)`

= `-e^(x-sin^2x)xx(1-2sinx xxcosx)`

= `-(1-sin2x)e^(x-sin^2x)`