derivative evaluate

**ocmisssunshine** · October 22nd 2007, 02:23 PM

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.

**TD!** · October 22nd 2007, 02:26 PM

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.

**topsquark** · October 22nd 2007, 02:27 PM

Originally Posted by ocmisssunshine

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 $f(t) = t - 1$ . 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: $f(x) \times g(x) = (fg)(x)$ .

So $\frac{d}{dt}(f^2(t))$ is done via the chain rule. Does that suffice? (Hint: the answer will be in terms of f and its derivative.)

-Dan

**ocmisssunshine** · October 22nd 2007, 02:27 PM

Its a graph but I cant figure out how to find it form just the graph.

**topsquark** · October 22nd 2007, 02:32 PM

Originally Posted by ocmisssunshine

Its a graph but I cant figure out how to find it form just the graph.

You are saying you have the graph of y = f(t)?

You may need to post it for us to help you better.

-Dan

**ocmisssunshine** · October 22nd 2007, 02:40 PM

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.

**Plato** · October 22nd 2007, 02:42 PM

Originally Posted by ocmisssunshine

Its a graph but I cant figure out how to find it form just the graph.

With most graphing utilities one needs to enter $[f(x)]^2$ .
Example: $\sin ^2 (x) = \left[ {\sin (x)} \right]^2$

**topsquark** · October 22nd 2007, 03:22 PM

Originally Posted by ocmisssunshine

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 $2x^2$ . That's easy, just use the power rule: $4x$ .

Now say you want to take the derivative of a function of y where y(x) is a function of x: $2y^2$ . Now we need to include in this the derivative of y in the following manner: $4y \cdot \frac{dy}{dx}$ . 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
$(f^2(x)+g^3(x))$ <-- We'll ignore the x = 2 for now.

Take the derivative of each term separately:
$f^2(x)$ has a derivative of $2f(x) \cdot \frac{df}{dx}$ .
(Note the similarities with the previous example with the y in it.)

$g^3(x)$ has a derivative of $3g^2(x) \cdot \frac{dg}{dx}$ .

So the derivative of the whole expression is
$2f(x) \cdot \frac{df}{dx} + 3g^2(x) \cdot \frac{dg}{dx}$

Now apply the x = 2 condition.

-Dan

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