# Thread: Derivative chain rule help

1. ## Derivative chain rule help

Basically the Problem gives a table, but the way it words the problem kinda confuses me a bit:
Suppose that the function f and g and their derivatives with respect to x have the following values x=0 and x=1
x f(x) g(x) f(x)prime g(x)prime
0 1 1 5 1/3
1 3 -4 -1/3 -8/3
(Sorry for the horribly made table)

Evaluate the derivatives with respect to x of the following combinations at given value of x:
a. 5f(x)-g(x), x=1
b. f(x)g^3(x), x=0
c. f(x)/(g(x)+1), x=1
d. f(g(x)), x=0
e. g(f(x)), x=0
f. (g(x)+f(x))^-2, x=1
g. f(x+g(x)), x=0

Anyways you don't need to do all of them for me, like just explain how to do the first 2 would be enough, thanks.

Basically the Problem gives a table, but the way it words the problem kinda confuses me a bit:
Suppose that the function f and g and their derivatives with respect to x have the following values x=0 and x=1
x f(x) g(x) f(x)prime g(x)prime
0 1 1 5 1/3
1 3 -4 -1/3 -8/3
(Sorry for the horribly made table)

Evaluate the derivatives with respect to x of the following combinations at given value of x:
a. 5f(x)-g(x), x=1

$\displaystyle (5f(1)-g(1))' = 5f(1)'-g(1)' = 5\times \frac{-1}{3}-\frac{-8}{3} = \dots$

3. For the first two, we have

$\displaystyle \frac{d}{dx}(5f(x)-g(x))=5\frac{d}{dx}f(x)+\frac{d}{dx}g(x)=5f'(x)+g' (x).$

\displaystyle \begin{aligned} \frac{d}{dx}(f(x)g^3(x))&=\left(\frac{d}{dx}f(x)\r ight)g^3(x)+f(x)\left(\frac{d}{dx}g^3(x)\right)\;\ ;\;\;\;\;\;\;\;\;\mbox{(Product Rule)}\\ &=f'(x)g^3(x)+f(x)(3g^2(x)\cdot g'(x)).\;\;\;\;\;\;\;\;\;\;\mbox{(Power Rule, Chain Rule)} \end{aligned}

4. Thats what I thought at first, but my friend said that u had to apply the chain rule?

5. You do have to apply the chain rule in some of these questions.

For d) and e) you need to.