If #f(x)= 9sin(sinx^2)#, then what is #f^'(x)#?

1 Answer
Jul 29, 2015

Answer:

#f^'(x) = 18x * cosx^2 *cos(sinx^2)#

Explanation:

Start by breaking down your function - this will give you an idea about what you need to use to differentiate it

#f(x) = y = 9 * sin(sin(x^2))#

From the looks of it, you're going to need to use the chain rule twice to differentiate #sin(sin(x^2))#.

The chain rule tells you that you can differentiate a function that is actually the composition of two other functions like this

#color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))#, where

#y# is a function that depends on variable #u#, which in turn depends on variable #x#.

Another important thing to remember is that

#d/dx(sinx) = cosx#

So, you have #sin (sinx^2)# as part of your original function. First, take #sin(x^2)# to be #u# so that you have

#y = 9 * sinu#

This can be differentiated like this

#d/dx(y) = d/(du)(9 * sinu) * d/dx(u)#

#d/dx(y) = 9 * d/(du)(sinu) * d/dx(u)#

Now you need to use the chain rule for a second time, since you have

#u = sinx^2#

This time, you can write #u = sinu_1#, where #u_1 = x^2#. This will get you

#d/dxu = d/(du_1) * sinu_1 * d/(dx)u_1#

#d/dx(u) = cosu_1 * 2x#, which is equivalent to

#d/dx(u) = cosx^2 * 2x#

Your original derivative now becomes

#d/dx(y) = 9 * cos(sinx^2) * 2x * cosx^2#

which is equal to

#d/dx(y) = color(green)(18x * cosx^2 * cos(sinx^2))#