Asked 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?
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?
Answers
Answered by
Count Iblis
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.
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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