# Finding derivatives based on the graph

Use the graph below to complete the following where h(x) = f(x)g(x) and m(x) = f(x)/g(x)

1. h'(x)
2. m'(x)
3. m'(2)
4. h'(−7)
5. h'(-5)

The first two problems are quite straightforward since we know that h(x) = f(x)g(x) and m(x) = f(x)/g(x).

1. h'(x) = f'(x)g(x) + f(x)g'(x)

2. m'(x) = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2

Starting from number 3, I am not sure how I am supposed to do it. I know that by plugging-in the value, we get:

3. m'(2) = (f'(2)g(2) - f(2)g'(2)) / (g(2))^2

where g(2) = -1 and f(2) = 1.3 (not sure about f(2))

In this problem, how am I supposed to find f'(2) and g'(2)? I can't quite figure it out. Any help would be appreciated. Thanks!

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Notice that $\ f(x) \$ is a straight line (thus, constant slope) on the interval $\ 0 \ \le \ x \ \le \ 6 \$ and $\ g(x) \$ has a horizontal tangent at $\ x \ = \ 2 \ .$ Find the slope of the line for $\ f(x) \$ and you'll have all the information you need. Something similar will work for #4 ; beware on #5 : what can be said about $\ f'(-5) \$ ? –  RecklessReckoner Jan 12 at 22:10
@RecklessReckoner In this case, f'(2) = -4/3 ((y2- y1) / (x2 - x1)) and g'(2) = 0. Then the answer for #3 will be m'(2) = DNE because we get the expression divided by 0. Is this correct? For #5, h'(-5) = DNE because f'(-5) = DNE (sharp corner). –  KurodaTsubasa Jan 12 at 22:38
Yes on #5: $\ f(x) \$ has no defined derivative at $\ x \ = \ -5 \ ,$ so neither will $\ f(x) \$ . On #3 , it's $\ g'(x) \$ that is zero at $\ x = 2 \$ , but $\ g(2) \$ is not zero (as you already noted)... –  RecklessReckoner Jan 12 at 22:44
@RecklessReckoner Oh, yes, my mistake. I mean the answer would be m'(2) = 4/3. Now I understand how I should approach these questions. Thank you very much! –  KurodaTsubasa Jan 12 at 23:05
Glad to hear that: this is a sort of question that sometimes appears on exams, using a graph, or tabular data, or specified functions. –  RecklessReckoner Jan 12 at 23:08