Im confused on what this is telling me.
d/dt (f^2(t)) at t=2 i know I have to find the derivative but how with the f^2 Ive never seen that before could someone please explain.
Say, for example $\displaystyle f(t) = t - 1$. 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: $\displaystyle f(x) \times g(x) = (fg)(x)$.
So $\displaystyle \frac{d}{dt}(f^2(t))$ 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 $\displaystyle 2x^2$. That's easy, just use the power rule: $\displaystyle 4x$.
Now say you want to take the derivative of a function of y where y(x) is a function of x: $\displaystyle 2y^2$. Now we need to include in this the derivative of y in the following manner: $\displaystyle 4y \cdot \frac{dy}{dx}$. 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
$\displaystyle (f^2(x)+g^3(x))$<-- We'll ignore the x = 2 for now.
Take the derivative of each term separately:
$\displaystyle f^2(x)$ has a derivative of $\displaystyle 2f(x) \cdot \frac{df}{dx}$.
(Note the similarities with the previous example with the y in it.)
$\displaystyle g^3(x)$ has a derivative of $\displaystyle 3g^2(x) \cdot \frac{dg}{dx}$.
So the derivative of the whole expression is
$\displaystyle 2f(x) \cdot \frac{df}{dx} + 3g^2(x) \cdot \frac{dg}{dx}$
Now apply the x = 2 condition.
-Dan