Question

graph help
f(x) and g(x)
x=3
f(x)=2
g(x)=-4
g'(x)=1/5

h(x)=2g(x)+1/(x)-f(x)g(x)

Answer 1

compute h'(3)=

Answer 2

help please @wio

Answer 3

\[\large h(x)=\frac{2g(x)+1}{x-f(x)g(x)}\]Is this what h(x) loos like?

Answer 4

yes sure is

Answer 5

So to find \(\large h'(x)\) we'll have to apply the quotient rule.

Answer 6

ok, g(x)f(x)'-f(x)g(x)'/g(x)^2?

Answer 7

It's probably not a good idea to think of the definition of the quotient rule in terms of \(\large f\) and \(\large g\) since that's going to confuse you in this problem :) lol

Because in this problem our \(\large f\) is \(\large 2g(x)+1\).
While our \(\large g\) is \(\large x-f(x)g(x)\).

Understand? :x

Answer 8

so far yes,

Answer 9

\[\large h(x)=\frac{u}{v}\]\[\large h'(x)=\frac{\color{royalblue}{u'}v-u\color{royalblue}{v'}}{v^2}\]

\[\large h'(x)=\frac{\color{royalblue}{\left[2g(x)+1\right]'}\left[x-f(x)g(x)\right]-\left[2g(x)+1\right]\color{royalblue}{\left[x-f(x)g(x)\right]'}}{\left[x-f(x)g(x)\right]^2}\]

So here is our setup for quotient rule.

Answer 10

It's a doozy. I tried to make it a little easier to read by changing the color of the ones we'll need to take a derivative of.

Answer 11

its really easy to see when you colored it thank you.

Answer 12

what is the next step?

Answer 13

So all we've done so far is setup the quotient rule. We now need to take the derivative of each blue part. Let's try to do it piece by piece.

Answer 14

So here is the upper left of our fraction.\[\large \color{royalblue}{\left[2g(x)+1\right]'}\left[x-f(x)g(x)\right]\]Taking the derivative gives us,\[\large \color{royalblue}{\left[2g'(x)+0\right]}\left[x-f(x)g(x)\right]\]

Understand how I did that?
We don't actually know what g(x) is, so we just throw a prime on it, to show that we've taken its derivative.

Answer 15

Yes one is constant. so it became zero? please help me solve this question. Its my homework that is due tomorrow

Answer 16

I want to get used to solve this question

Answer 17

So the upper left of our fraction will simplify to,\[\large \left[2g'(x)\right]\left[x-f(x)g(x)\right]\]

Now looking at the upper right of our fraction, this one is going to be a little tricky. We have to apply the product rule to the f and g part.
\[\large -\left[2g(x)+1\right]\color{royalblue}{\left[x-f(x)g(x)\right]'}\]

\[\large -\left[2g(x)+1\right]\color{royalblue}{\left[1-\color{orangered}{\left[f'(x)g(x)+f(x)g'(x)\right]}\right]}\]The x gave us 1, the product rule on f and g is in orange.

Answer 18

The brackets around the orange are really important since there was a negative in front of the f(x)g(x), it will need to be distributed through.

Answer 19

So far i understand. you explain very clearly

Answer 20

So we've taken all the derivatives that we needed to take. This gives us an h'(x) of,
\[\large \frac{\left[2g'(x)\right]\left[x-f(x)g(x)\right]-\left[2g(x)+1\right]\left[1-f'(x)g(x)-f(x)g'(x)\right]'}{\left[x-f(x)g(x)\right]^2}\]

Answer 21

Quite the ugly looking problem.. wow your teacher is mean :d

Answer 22

They want us to evaluate h'(x) at x=3.
Which will look like this,\[\large \frac{\left[2g'(\color{#CC0033}{3})\right]\left[x-f(\color{#CC0033}{3})g(\color{#CC0033}{3})\right]-\left[2g(\color{#CC0033}{3})+1\right]\left[1-f'(\color{#CC0033}{3})g(\color{#CC0033}{3})-f(\color{#CC0033}{3})g'(\color{#CC0033}{3})\right]'}{\left[\color{#CC0033}{3}-f(\color{#CC0033}{3})g(\color{#CC0033}{3})\right]^2}\]

And they gave us a bunch of info to fill in the blanks.
\[\large f(3)=2\]\[\large g(3)=-4\]\[\large g'(3)=1/5\]

Answer 23

But we have a small problem. In order to solve this, we need to know what \(\large f'(3)\) is.
Did you forget to paste some of the info?

Answer 24

haha really!
um there is a graph like this:
x f'(x) g(x) g'(x)
3 2 -2/3 1/5

Answer 25

it says compute h'(3)=

Answer 26

Oh so that was a small typo on your part I guess c: \(\large f'(3)=2\).
But then we still need \(\large f(3)\) grrr -_-

We can simplify this pretty far down, we won't be able to get a numerical answer though, no big deal.

Answer 27

Oh I forgot to erase the prime on the upper right fraction, don't let that confuse you please :O That should be gone.

Answer 28

hmm? which part?

Answer 29

\[\large \frac{\left[2g'(x)\right]\left[x-f(x)g(x)\right]-\left[2g(x)+1\right]\left[1-f'(x)g(x)-f(x)g'(x)\right]\color{red}{'}}{\left[x-f(x)g(x)\right]^2}\]
The one in red here. We have taken the derivative of everything inside. So the prime on the outside should disappear.

Answer 30

so far im with you

Answer 31

Grr it got cut off, imma have to paste it smaller.

Answer 32

\[\frac{\left[2(1/5)\right]\left[3-f(3)(-2/3)\right]-\left[2(-2/3)+1\right]\left[1-(2)(-2/3)-f(3)(1/5)\right]}{\left[3-f(3)(-2/3)\right]^2}\]

Answer 33

Thank you,

Answer 34

i can see the whole thing now :)

Answer 35

If you're confused about what I did, just look at the first set of brackets, maybe that will help you understand.

\(\large \left[2g'(3)\right]\) became \(\large \left[2(1/5)\right]\)

Because \(\large g'(3)=1/5\)

Answer 36

See the problem we have? We have a big ugly mess since we weren't given information about f(3).

Answer 37

yes I got it.:)

Answer 38

Otherwise we would be able to simplify it down to just a number.

Answer 39

Oh well :c heh

Answer 40

Cool thanks for helping me.

Answer 41

i have one question, may i ask?

Answer 42

Yes, if it's not another 30 minute problem c: lolol

Answer 43

yes, Thank you! its
\[w=\frac{ \sqrt(t) }{ 2t^2+t3^t }\]

find dw / dt

Answer 44

\[\large w=\frac{\sqrt t}{2t^2+t3^t}\]

Before taking this derivative, let's do some side work.
Do you know the derivative of this term?\[\large 3^t\]

Answer 45

something to do with In?

Answer 46

lol, yes something to do with natural log :)
We don't have time to go through the steps of that separately.
So i'll just remind you of the rule.

\[\large f(x)=e^x\]\[\large f'(x)=e^x\]Anytime the base is NOT e, you have to multiply by the natural log of the base.\[\large g(x)=2^x\]\[\large g'(x)=2^x(\ln 2)\]

Just keep that in mind for when it comes up in this problem.

Answer 47

\[\large w=\frac{\sqrt t}{2t^2+t3^t}\]
\[\large w'=\frac{\color{royalblue}{(\sqrt t)'}\left[2t^2+t3^t\right]-\sqrt t\color{royalblue}{\left[2t^2+t3^t\right]'}}{\left[2t^2+t3^t\right]^2}\]

Again, quotient rule.
And all we really need to worry about is the blue parts.

Answer 48

You lied! This is an evil problem -_- just like the last one. lolol

Answer 49

I am so sorry...

Answer 50

I cant even tell weather its not evil or not since I have no idea

Answer 51

i mean not evil or evil

Answer 52

If you don't have the derivative of sqrt memorized, then you'll want to change it to a rational expression so you can apply the power rule.

Answer 53

Do you know that derivative? c: It's a good one to remember.

Answer 54

power rule? n-1? and bring one number front

Answer 55

yes... but do you understand how to take the derivative of the square root? :o

Answer 56

No...

Answer 57

\[\large \sqrt t \qquad = \qquad t^{1/2}\]You can take the derivative from this point by applying the power rule.

Answer 58

oh i see. so it should 1/2t^(-1/2)

Answer 59

ya

Answer 60

ok, could you write the whole thing?

Answer 61

please dont go @zepdrix

Answer 62

write the whole thing? :3 pshhhh ill bet you can do that XD
Let's look at the other blue part a moment.

We'll have to apply the product rule again.

Answer 63

ok,i have been struggling with this problem for 1hr and half.

Answer 64

\[\large \left[2t^2+t3^t\right]' \qquad = \qquad \left[4t+(t)'3^t+t(3^t)'\right]\]So here what i've done is setup the product rule, we'll still need to evaluate it.

Answer 65

yes this is what i am struggling with!

Answer 66

\[\large \left[4t+1\cdot3^t+t3^t(\ln 3)\right]\]Understand how that worked out?

Answer 67

yes, so far. you used the rule that you explained before right

Answer 68

ya

Answer 69

ok, whats next step

Answer 70

So we were taking the derivative of each piece separately. Now just put the pieces back in.
\[\large \frac{1/2t^{-1/2}\left[2t^2+t3^t\right]-\sqrt t\left[4t+1\cdot3^t+t3^t(\ln 3)\right]}{\left[2t^2+t3^t\right]^2}\]

You could probably simplify it a bit.. but no big deal..

Answer 71

how you simpligy Sqrt(t) and In3 part?

Answer 72

i would appreciate it if you could show me how to simplify the whole thing

Answer 73

Generally with these types of problems its better to not simplify them too far unless your teacher asks you to.
They usually just want to see that you know how to take the derivative.
It's not worth going any further really. :o

Answer 74

Ok, thank you.

Answer 75

you are the best math teacher :)

Answer 76

lol haf good night c:

Answer 77

thank you for your time night night!