# Homework Help: Calculate derivatives

1. Oct 7, 2007

### physstudent1

1. The problem statement, all variables and given/known data
Suppose f and g are functions. Assume that f is differentiable and f(1) = 2, f'(1)=4 and that g(x) = x^4 -x+1. Use this information to calculate:

(fg)'(1)
(f o g)'(1)

2. Relevant equations

3. The attempt at a solution

I'm unsure of what this wants me to do Does it want me to multiply the derivatives of f and g together and calculate it at 1 or does it want me to multiply f and g then find the derivative and plug in 1.

For the 2nd one I found the derivative of g(x) to be 4x^3 -1 plugged a 1 into that and got 3 now i have f'(3) = ? how can i know that when they only gave me f(1) and f'(1) ?

Last edited: Oct 7, 2007
2. Oct 7, 2007

### Avodyne

I would guess that (fg)(x) means f(x) times g(x), but you should check your book or course notes or whatever is relevant for the definition you are supposed to be using.

For the second part, (f o g)(x) usually means f(g(x)), which can be differentiated by the chain rule. You should find that you do not need the value of f'(3).

3. Oct 7, 2007

### atqamar

Does "product rule" and "chain rule" ring a bell?
Product rule: $$(fg)'=f'g+fg'$$
and
Chain rule: $$(f \circ g)'(x) = f'(g(x)) g'(x)$$

Just a clarification of notation: $$(f \circ g)(x)$$ is also written as $$f(g(x))$$ and is read as "$$f$$ of $$g$$ of $$x$$". It simply means you first apply the $$g$$ function to $$x$$ and then the $$f$$ function to the resultant. The $$'$$ indicates the first derivitive of the function.

4. Oct 7, 2007

### physstudent1

ah ok its not (fg)x tho its (fg)'(x)

and I don't know the equation for f(x) it doesn't give it to you so how can i find the derivative of

f(x^4-x+1)
would it be
(3x^3 - 1)*(f(x^4 - x+1))

5. Oct 7, 2007

### atqamar

You don't need the equation for $$f(x)$$.

Ok, lets do the first one first. Start off with the product rule:
$$(fg)'=f'g+f'$$ which can be written more formally as:
$$(f(x)g(x))'=f'(x)g(x) + f(x)g'(x)$$

The values of $$f'(1)$$ and $$f(1)$$ are given, and the values for $$g'(1)$$ and $$g(1)$$ can be calculated by plugging $$1$$ into the $$g(x)$$ function. And then differentiate $$g(x)$$ to find $$g'(1)$$. Plug in the resulting values into the above equation and you will obtain your solution for the first part.

Finding the derivitive of $$g(x)$$ should be quite simple.

To do the second part, similarly just plug in your knowns.

Last edited: Oct 7, 2007
6. Oct 7, 2007

### physstudent1

gotcha thanks