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

1 Answer
Mar 19, 2017

Answer:

#(df)/(dx)=-1/2(1-sin2x)sqrt(e^((x-sin^2x)))#

Explanation:

Chain Rule - 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)#.

Here for example, we have #y=-sqrt(e^(x-sin^2x))#

We can also write this as #y=-e^((x-sin^2x)/2)#

and we can say #y=f(x)=-e^(g(x))#, where #g(x)=(x-sin^2x)/2#

and #(df)/(dx)=-e^(g(x))xx(dg)/(dx)#

= #-e^(g(x))xx(1/2-1/2xx2sinx xx cosx)#

Observe that to differentiate #sin^2x# we have again used chain rule as differential of #sin^2x# w.r.t #sinx# is #2sinx# and then multiplied it by #d/dx sinx=cosx#

Hence #(df)/(dx)=-e^((x-sin^2x)/2)xx1/2(1-sin2x)#

= #-1/2(1-sin2x)sqrt(e^((x-sin^2x)))#