Thread: Chain Rule

Hello, I have been gearing up for Calculus 1 after being out of school for ~13 years. I've been using the Math Tutor DVD for Calculus 1 to get myself ready before class starts and it has been good thus far. Was not too comfortable with my chain rule skills from the examples given on the video so found some online to practice. The following problem has given me grief:
f(x)=sqrt(5cosx)
I solve it this way:
f(x)=(5cosx)^(1/2)
f'(x)=1/2(5cosx)^(-1/2) * (-sinx)
f'(x)=-(sinx)/2sqrt(5cosx)

The solution on the webpage gives:
f'(x)=1/2(5cosx)^-1/2 * 5(-sinx)

My question is how did that "5" make it to 5(-sinx)??
Shouldn't the derivative of (5cosx) be (-sinx) since 5 is a constant and is eliminated?

Originally Posted by aorelup

Hello, I have been gearing up for Calculus 1 after being out of school for ~13 years. I've been using the Math Tutor DVD for Calculus 1 to get myself ready before class starts and it has been good thus far. Was not too comfortable with my chain rule skills from the examples given on the video so found some online to practice. The following problem has given me grief:
f(x)=sqrt(5cosx)
I solve it this way:
f(x)=(5cosx)^(1/2)
f'(x)=1/2(5cosx)^(-1/2) * (-sinx)
f'(x)=-(sinx)/2sqrt(5cosx)

The solution on the webpage gives:
f'(x)=1/2(5cosx)^-1/2 * 5(-sinx)

My question is how did that "5" make it to 5(-sinx)??
Shouldn't the derivative of (5cosx) be (-sinx) since 5 is a constant and is eliminated?

Not quite!

5cosx is cosx+cosx+cosx+cosx+cosx.

Now try differentiating it again!

Thank you very much. Makes perfect sense when viewed that way.