posted by Marissa .
Let g(x) = sin (cos x^3) Find g ' (x):
The choices are
a) -3x^2sinx^3cos(cos x^3)
b) -3x^2sinx^3sin(cos x^3)
c) -3x^2cosx^3sin(cos x^3)
d) 3x^2sin^2(cos x^3)
I'm not exactly sure where I should start.
Should I begin with d/dx of sin? Or do the inside derivative first...and do I have to separate cos and x^3 as well?
This is a multiple choice question and then the way you should attack the problem should be different than if you were asked to find the derivative of sin[cos(x^3)]
What you do is you use the chain rule, accrding to which the derivative of
sin(f(x)) = cos(f(x)) f'(x)
Without calculkating anything, you immediately see that b) c) and d) cannot be right, so a) must be right.
The prefactor -3x^2sin(x^3) is indeed the derivative of the argument of the sin. You only have to notice that it looks correct and then move on to the next question. In a test you can be given many multiple choice question and then the teacher will test if you can spot the correct answer within seconds.