It looks like there's a given function f, and you are asked to find the derivative of its square, f², with respect to t in t = 2.
Say, for example . Then the function [tex]f^2(t) = (t - 1) \times (t - 1) = (t - 1)^2[tex].
This is an extension of the notation for multiplying two functions: .
So is done via the chain rule. Does that suffice? (Hint: the answer will be in terms of f and its derivative.)
-Dan
I will try that one on my own cause I cant upload the graph but can you help me with this instead.... this is given...
f(2)= 4
f '(2)= 4
f ''(2)=-1
g(2)=2
g '(2)=5
I have to find the derivative of (f^2(x)+g^3(x)) at x=2 I cant understand the chain rule for the life of me.
The chain rule is a way to take the derivative of a function of a function.
Say, for example, you want to take the derivative of . That's easy, just use the power rule: .
Now say you want to take the derivative of a function of y where y(x) is a function of x: . Now we need to include in this the derivative of y in the following manner: . The first factor is just like the x case; all we've added is the derivative on the end.
So let's take a look at your problem:
Take the derivative of
<-- We'll ignore the x = 2 for now.
Take the derivative of each term separately:
has a derivative of .
(Note the similarities with the previous example with the y in it.)
has a derivative of .
So the derivative of the whole expression is
Now apply the x = 2 condition.
-Dan