Question

Help with Chain Rule using a table to find the derivatives?

Answer 1

Sure, do you have a specific example?

Answer 2

Honestly, this looks like more of a pain than just simply calculating the derivatives outright, but... We'll do it with the table anyways.

So for the first one, it says h'(3). So first off, what is h'(x) if we want to know that we need to write it in terms of f and g and their derivatives. So looking at the top left, it says h(x)=f(g(x)).

So then can you find the derivative of f(g(x)) using the chain rule? Give it your best guess and I'll help you along the way so you can do it yourself when you get to the test! =D

Answer 3

I know that the chain rule looks like f'(g(x))*g(x) so I guess h'(3)= f'(g(3))*g'(3) right?

Answer 4

Yeah, exactly! Can you take it further by looking at the table now?

Answer 5

g(3)=1 so I replace it and get f'(1) which is 5 and g'(3) is 20

Answer 6

Yep, almost done, perfect.

Answer 7

5*20=100 so h'(3)=100

Answer 8

Perfect, you got the answer all on your own. =D

Answer 9

I can help check your answers if you keep going, just post them as you get them if you want. I might not be immediate since I'm working on doing some homework right now too.

Answer 10

Now I get the first 4 ones, but I'm really confused by the last 2

Answer 11

Sure, so show me your steps and where you get stuck, I'll see what I can do.

Answer 12

For q'(0) it should look like g'(f(e^2x+1))*f'(e^2x+1) right?

Answer 13

Almost, except when you take the derivative, it's a chain rule inside of a chain rule. So when you take the derivative of g, you have to multiply it by the derivative of the inside. So that's the derivative of f. But since f has something inside of it, when you take the derivative of the outside of that, you need to multiply that by the derivative of the inside as well. I hope that makes sense.

Answer 14

I kind of get what you're saying but could you show me please?

Answer 15

Sure, so when we take the derivative of g(f(e^2x+1), lets just call e^2x+1=m(x).

Then we have g(f(m(x))). The derivative will be:

g'(f(m(x)))*f'(m(x))*m'(x)

Now we can plug m(x) back in, and notice we have m'(x), so we can calculate that as: m'(x)=2e^2x, no problem.

g'(f(e^2x+1))*f'(e^2x+1)*(2e^2x)

If that doesn't help show the steps, consider just taking the derivative of that inside part, don't even think about the g(...) part.

The derivative of f(e^2x+1) is going to be f'(e^2x+1)*(2e^2x) right? Maybe think about it a second and it'll make sense, it's all sort of abstract I'm sorry!

Answer 16

I got q'(0)=60 but the answer in the book is 120...?