Question

find h' (x), where f(x) is an unspecified differentiable function. Use the notation f' to denote the derivative of f
Example : if h(x)=4(x)^2 then h'(x)=4(2f(x)f'(x))=8f(x)f'(x)
a) h(x)=3f(x)^4
B) h(x)=6x^8f(x)
C)h(x)=(1-e^x)^3/f(x)
D) H(x)=5sqrtf(x)^4

Answer 1

\[\Large\bf\sf h(x)\quad=\quad 3f(x)^4\]Mmmm so what are you having trouble with?

Answer 2

Function notation can be hard to get comfortable with.
f(x) just means that the right side contains a bunch of `stuff` involving x.
They didn't specify exactly what type of stuff.

But we can apply the rules that we know.

Answer 3

\[\Large\bf\sf h(x)\quad=\quad 3(stuff)^4\]\[\Large\bf\sf h'(x)\quad=\quad 3\cdot 4(stuff)^3(stuff)'\]Power rule, then chain rule, yes?

Answer 4

\[\Large\bf\sf h(x)\quad=\quad 3f(x)^4\]\[\Large\bf\sf h'(x)\quad=\quad 3\cdot4f(x)^3\cdot f'(x)\]

Answer 5

Where you at Louis!! :L

Answer 6

confused:(

Answer 7

do i plug in h(x)=3f(x)^4 into H'

Answer 8

The second line that I wrote, that derivative,
that one is done.

Part a) is done.
You don't go any further than that.
(You can multiply the 3 and 4 together I guess).

Answer 9

Are you confused how I got that line?

Answer 10

hold on

Answer 11

so it will be \[12f(x)f'(x)\]

Answer 12

Yes.

Answer 13

ok i was reading it wrong

Answer 14

Ah :3

Answer 15

Understand how to do the next one?
Looks like we have the `product` of two x-things.

Answer 16

no not really but i'm trying to work it out now

Answer 17

If you have a `product`, you use a certain `ruuuuuuule` :X

Answer 18

will it start of 8x^6

Answer 19

?? 0_o

Why did your 8 and 6 switch places?

Answer 20

because i used\[d/dx (u^n)\]

Answer 21

=nu^n-1

Answer 22

\[\Large\bf\sf \frac{d}{dx}(x^n)\quad=\quad nx^{n-1}\]So the 8 comes down, good.
Then your 8(exponent) decreases to a 7... yes?

Answer 23

or i think it should be 48x^7

Answer 24

yah that looks better.

Answer 25

ok

Answer 26

\[\Large\bf\sf h(x)=6x^8f(x)\]
\[\Large\bf\sf h'(x)\quad=\quad \color{royalblue}{\left[6x^8\right]'}f(x)+6x^8\color{royalblue}{\left[f(x)\right]'}\]Did you remember about product rule?
Here is what the setup should look like.
Where the blue parts are the derivatives you need to take.

Answer 27

\[6x^ 8f(x)+48x^7f(x)\]

Answer 28

\[48x^7f(x)+6x^8f(x)\]

Answer 29

Take the derivative of the second f.
Your result looks close,\[48x^7f(x)+6x^8f'(x)\]

Answer 30

\[48^7f(x)+6x^8f\]

Answer 31

? :\

Answer 32

\[48x^7f(x)+6x^8f\]

Answer 33

i don't really understand how to take the derivative of the second f

Answer 34

Oh boy...

The f is NOT a constant coefficient multiplying an x.
It's a function of x.

The derivative of f(x) is f'(x).
You stop there....

We don't KNOW what f(x) is, we can't go any further in defining its derivative.

Answer 35

so if the f'(x) is constant then it will be \[48x^7f(x)+6x^8\]

Answer 36

If for some reason they had told us that f'(x) was constant, then we could write it as:\[48x^7f(x)+6x^8\cdot C\]Where C is our constant derivative.
They didn't tell us that though :o

Answer 37

ok so your saying that the f(x) derivative is f'(x), so therefor my answer should be \[48x^7f(x)+6x^8f'(x)\]

Answer 38

yes.

Answer 39

ok i put that at first but you told me to look at the problem again:(

Answer 40

oh no i didn't sorry

Answer 41

ya yours was close :c maybe was just a typo before.

Answer 42

ok so now i have to plug it into h'(x) right

Answer 43

No. You have finished part b).\[\Large\bf\sf h(x)\quad=\quad 6x^8\;f(x)\]\[\Large\bf\sf h'(x)\quad=\quad 48x^7\;f(x)+6x^8\;f'(x)\]

Answer 44

Unless that's what you meant by plug it in.. :o

Answer 45

ok

Answer 46

\[\Large\bf\sf h(x)=\frac{(1-e^x)^3}{f(x)}\]Oh boy, here's where they start to get tricky :D

Answer 47

lol....ok i'm about to try

Answer 48

ok can you tell me if i'm on the right track?
\[f(x)3(1-e^x)^2-(1-e^x)^3f'(x)/(f(x))^2\]

Answer 49

Yesssss nice job, you're very close.
Just need to fix this orange part a sec,\[\Large\bf\sf h'(x)\quad=\quad \frac{f(x)\cdot\color{orangered}{3(1-e^x)^2}-(1-e^x)^3\;f'(x)}{f(x)^2}\]

Answer 50

Need to apply the chain rule, not just the power rule to the outside.

Answer 51

Multiply by the derivative of the inside.

Answer 52

\[\frac{ f(x)3(1-e^2x)^2-(1-e^x)^3f'(x) }{f(x)^2 }\]

Answer 53

\[\Large\bf\sf \frac{d}{dx}(1-e^x)^3\quad=\quad 3(1-e^x)^2\color{royalblue}{\frac{d}{dx}(1-e^x)}\]Chain rule^
This is what you should be getting for that orange part,\[\Large\bf\sf \color{orangered}{3(1-e^x)^2(-e^x)}\]

Answer 54

ok

so that will be \[\frac{ f(x)3(1-e^x)^2(-e)-(1-e^x)^2f'(x) }{ f(x) ^2}\]

Answer 55

Don't forget the x exponent on your e! :O
And you changed the second (1-e^x) term.. hmm it was correct before.
Should get something like this,\[\Large\bf\sf h'(x)\quad=\quad \frac{f(x)\cdot\color{orangered}{3(1-e^x)^2(-e^x)}-(1-e^x)^3\;f'(x)}{f(x)^2}\]

Answer 56

ok now i factor

Answer 57

I would ummmmm.. leave it like that.
You can maybe simplify it a little bit, but I wouldn't fuss with it too much.
Just bring the `negative` and `3` to the front.

Answer 58

ok

Answer 59

hello Zepdrix, can you help me with this last problem i'm stuck on

Answer 60

\[\Large\bf\sf H(x)=5\sqrt {f(x)^4}\]

Answer 61

Did I format that correctly?
The 4th power is inside of the root like that?

Answer 62

yes ok this is what I have so far\[1/5(f(x)^4)^-4/5*(f(x)^4)\]

Answer 63

the-4/5 is an exponent

Answer 64

Oh is it supposed to be a 5th root?
Is that what the 5 is about?

Answer 65

\[\Large\bf\sf H(x)=\sqrt[5]{f(x)^4}\]

Answer 66

\[\Large\bf\sf H(x)=f(x)^{4/5}\]

Answer 67

yes

Answer 68

\[\Large\bf\sf H'(x)=\frac{4}{5}f(x)^{-1/5}\cdot f'(x)\]

Answer 69

I'm a little confused by your coefficient and exponent, they look backwards. Hmm

Answer 70

ok I started off like this\[h(x)=(f(x)^4)^1/5\]

Answer 71

Hmm you should simplify it a little further before differentiating.
Looks like you're giving yourself more work D:
\[\Large\bf\sf \sqrt[5]{a^{4}}\quad=\quad (a^4)^{1/5}\quad=\quad a^{4/5}\]

Answer 72

Otherwise, from where you left off,\[1/5(f(x)^4)^{-4/5}*\color{orangered}{(f(x)^4)}\]You still need to chain rule this, applying the power rule,
then chain rule again to get your f'(x)

then a bunch of simplification.
It will get pretty messy :\

Answer 73

\[\frac{ 1 }{ 5}(f(x)^4)^{-4/5}(4f'(x))\]

Answer 74

then I put \[\frac{ 4f'(x) }{5 \sqrt[5]{f(x)^4} }\]

Answer 75

\[\frac{ 1 }{ 5}(f(x)^4)^{-4/5}\color{orangered}{(4f(x)^3)f'(x)}\]

Answer 76

You forgot your power rule.

Answer 77

This approach is so messyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

Do it the other way :(
Ugh... there is so much room for error doing it this way....

Answer 78

You're doing an extra chain,
and 2 extra simplification steps...

Answer 79

ok sorry but this is kind of the only way I know

Answer 80

\[\Large\bf\sf H(x)\quad=\quad \sqrt[5]{f(x)^4}\quad=\quad\left(f(x)^4\right)^{1/5}\quad=\quad \color{orangered}{f(x)^{4/5}}\]That's the only step you're missing.
Writing it as a rational expression (fraction for your exponent).\[\Large\bf\sf H(x)\quad=\quad f(x)^{4/5}\]

Answer 81

Your power rule and chain will work out nicer from there. :o

Answer 82

ok thankyou I understand now