Question

How do I solve this? f(x)=-2h(x)+(x/h(x)), h(-1)=-3, h'(-1)=-2. Find f'(-1).

Answer 1

Well, the problem says we're looking for f'(-1), which is f'(x) when x=-1.

It looks like we know f(x), so we should be able to find the f'(x in terms of h(x) and h'(x) ) by taking the derivative of both sides with respect to x.

Answer 2

Do I just plug in h'(x) into the f(x). Or if im wrong could you tell me how to get the derivative of f(x)

Answer 3

just a bit of implicit differentiation:
\( \displaystyle f(x) = -2 h(x) + \frac{x}{h(x)} \)
\( \displaystyle \frac{\text{d}}{\text{d}x} f(x) = \frac{\text{d}}{\text{d}x} \left( -2 h(x) + \frac{x}{h(x)} \right) \)

Where the regular properties of derivatives still apply: sum rule and quotient / product rule for the RHS second term... should work. Does that make sense?

Answer 4

Ohh so I just use \[f'(x)+g'(x)= d/dx(f(x)+g(x))\]

so then f'(-1)=-2(-3) + (-1/-3) correct?

Answer 5

where did you get that answer from?

Answer 6

\( \displaystyle \frac{\text{d}}{\text{d}x} \left( -2 h(x) + \frac{x}{h(x)} \right) = \frac{\text{d}}{\text{d}x} \left( -2 h(x) \right) + \frac{\text{d}}{\text{d}x} \left( \frac{x}{h(x)} \right) \)
\( \displaystyle = -2 \frac{\text{d}}{\text{d}x} h(x) + \frac{\text{d}}{\text{d}x} \left( \frac{x}{h(x)} \right) \)

we should have -2h'(x) for the first part. The 2nd term requires quotient rule:
\(\displaystyle \frac{d}{dx} \left( \frac{f}{g} \right) = \frac{ f'g - fg'}{g^2} \)

Answer 7

Ok I am starting to see what to do. Now we have to take the derivitive of x/h(x). So the derivitive is

\[\frac{(1)(h(x))-(x)(h'(x))}{ h(x)^{2} }\]

So I just solve the derivitive of the one above then add
\[-2\frac{ d }{ dx } h(x)\]

to get f'(x)


Correct me if Im wrong

Answer 8

Yep, that is correct. :)

Answer 9

So ultimatly it turns into

\[f'(x)\frac{ 2 }{ 3 }x+4 \]

Answer 10

sorry forgot the = sign after f'(x)

Answer 11

what was the work to get that? since f'(x) would be before we sub x=-1, we should still have those h'(x) and h(x) in there

Answer 12

Hmm I will double check

Answer 13

* and once we sub x=-1, we should actually have no x's in the f'(-1) expression. :)

Answer 14

Hmm I will double check

Answer 15

f(x) = -2 h(x) + x/h(x)

--> f'(x) = -2 h'(x) + (h(x) - xh'(x))/(h(x))^2
this is what we established earlier
i think it should make sense to sub x=-1 right here, since we will get h(-1) and h'(-1):
--> f'(-1) = -2 h'(-1) + (h(-1) - (-1) h'(-1))/(h(-1))^2
and we just have to sub in h(-1) = -3 and h'(-1) = -2 there

Answer 16

So I started off by plugging in:
\[(-2)\frac{ d }{ dx } h(x)= (-2)(-2)=+4\]

Then I took the derivative of:

\[\frac{ (1)(h(x))-(x)h(x)) }{ h(x)^2 } = \frac{ (1)(-3)-(x)(-2) }{ (-3)^2 }=\frac{ -3 +2x }{ 9 }\]

Answer 17

ohh alright, yeah, so you just missed plugging in x=-1 for that remaining x. :)

Answer 18

So I just plug in and I am all good! Awesome thank you so much for your help

Answer 19

Thanks for your patience to.

Answer 20

Yep. The summary is just:
- We identify we're looking for a derivative.
- Take derivative of both sides
- Use properties of derivatives to simplify down to h(x) and h'(x)
- Substitute x=-1 directly
- We should end up with the answer. :)

You're welcome!

Answer 21

Just to confirm, what do you end up with now? :)

Answer 22

31/9

Answer 23

Alright, yeah, that's correct. Glad to have helped! :D

Answer 24

Thank you so much and have a wonderful day! :D