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

1 Answer
Oct 9, 2016

Answer:

#(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)#