chain rule in calculus?

Question

I am really confused on how to do these problems. i know you use the chain rule but I am not getting the correct answers. thanks so much if you can help and expain how to do it:

1)
f(x) = ln(5sinx^2)
f ' (x)=?

2)
F(y) = yln(9 + e^y)
F '(y)=?

3)
y = (ln (3x))/x^3
y'(x)=?
y"(x)=?

4)
This one is find an equation of the tangent line to the curve at the point (4,0)
y=ln(x^2 - 15)
y=?

5)
y=(sin4x)^lnx
y'=?

Thanks so much for any help you can give, I would really appreciate it!!!!

Answer 1

Well, I suppose the easiest way to understand the chain rule is just to say that all you are doing is differentiating multiple times in the same step.

I don't want to solve all of the problems so I'll just take number two and explain the steps for it.

First, it's a product rule differentiation(nothing to do with the chain rule right now) to get ln(9+e^y) + y*d/dy[ln(9+e^y)].

To start, just differentiate the ln(9+e^y) like you normally would to get 1/(9+e^y), but since the inner expression also has the variable that is being differentiated, it also needs to be differentiated in the same step.
So, multiply the 1/(9+e^y) by the derivative of 9+e^y by itself, which is e^y. This would equal e^y/(9+e^y)

Adding in the product rule step from before, the final solution is ln(9+e^y) + (y*e^y)/(9+e^y).

Answer 2

It helps if you keep naming variables.

For example, on the 1st one, f = ln(y), where y = 5 sin(x^2)

So f' = 1/y * y'.

That's the chain rule.

But you still need to find y'. Well, y = 5sin(z), where z = x^2.

So y' = (5 cos z ) * z'

And of course z' = 2x.

Now just keep substituting your intermediate results back in until everything is in terms of x, and you'll be done.