Posted by **Marissa** on Sunday, May 11, 2008 at 6:27pm.

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?

- Math, derivatives -
**Count Iblis**, Sunday, May 11, 2008 at 6:44pm
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.

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