# Thread: How do I start this derivative problem?

1. ## How do I start this derivative problem?

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

The graph:

2. ## Re: How do I start this derivative problem?

Find each of the following what? You already showed the graph, so you should be able to figure out $f(x)$ and $g(x)$ as piecewise continuous functions.

$f(x) = \left\{\begin{array}{ll}m_1 x + b_1 & -1

$g(x) = \left\{\begin{array}{ll}m_3 x + b_3 & -1 < x < 0 \\ m_4 x + b_4 & 0 < x < 3 \\ m_5 x + b_5 & 3 < x < 6 \end{array} \right.$

Find the five slopes and five corresponding y-intercepts (to find the y-intercepts, you pretend that each line segment continues to figure out where it would hit the y-axis if it were a line). To be the most accurate, try to find specific points that the line segments cross. For example, the left-most red line segment passes between the points $(-1,5)$ and $(1,2)$, so you can use those points to find the slope of that line segment, which would tell you $m_1$.

What else do you need help figuring out?

3. ## Re: How do I start this derivative problem?

So how would I find u'(x) from u(x)? and v'(x) from v(x)?

4. ## Re: How do I start this derivative problem?

Originally Posted by rhcprule3
So how would I find u'(x) from u(x)? and v'(x) from v(x)?
There are at least two ways of finding those derivatives. One way: use the product rule to find $u'(x)$ and the quotient rule to find $v'(x)$. Another way: write out the piecewise functions and differentiate. For example:

$u(x) = \left\{ \begin{array}{ll} (m_1 x + b_1)(m_3 x + b_3) & -1 < x < 0 \\ (m_1 x + b_1)(m_4 x + b_4) & 0 < x < 2 \\ (m_2 x + b_2)(m_4 x + b_4) & 2 < x < 3 \\ (m_2 x + b_2)(m_5 x + b_5) & 3 < x < 6\end{array} \right.$

And similarly for $v(x)$.

5. ## Re: How do I start this derivative problem?

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?

6. ## Re: How do I start this derivative problem?

Originally Posted by rhcprule3
The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?
Now I am confused. Do you want u(1) or u'(1)?

Assuming you want u'(1), do you know the product rule? If you do, then find u'(x) and plug in 1.

7. ## Re: How do I start this derivative problem?

Originally Posted by rhcprule3
The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

u(1) =

v(1) =

this is the actual question

so to find u'(1), I would need to use the product rule?
v(x)=f(x)g(x)
u(x)=f(x)g(x)

What is the distinction between u and v? Are you sure this is product problem or it is a composition problem, that is
u(x) = f(x) o g(x)
v(x) = g(x) o f(x)

8. ## Re: How do I start this derivative problem?

If I read it correctly (and there are some strange characters in the post), u= f times g and v= f divided by g.

Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.

9. ## Re: How do I start this derivative problem?

Originally Posted by HallsofIvy
If I read it correctly (and there are some strange characters in the post), u= f times g and v= f divided by g.

Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.
I went backward to the first post, you got me, it looks like a division OP rewrite it as product in the fifth post, probably a typo. That confused me.