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

1 Answer
Oct 9, 2016

(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)