Question

Find the Derivative, Using Chain Rule.
g(x)=9/5x^2

Answer 1

Hey Phi!\[\Large\rm g(x)=\frac{9}{5x^2}\]Is this what the function looks like?
With the x^2 in the bottom?

Answer 2

Yes :D

Answer 3

Changing some things up before finding a derivative,
\[\Large\rm =\frac{9x^{-2}}{5}=\frac{9}{5}x^{-2}\]Ok with those steps?
This form allows us to easily apply our power rule.

Answer 4

Imma go make some foods :3
You think about it!

Answer 5

Okay, so i know you cant have an exponet in the demoninator which is why you brought it up.
then you should multiply the -2 to the constant
so it would be
\[\frac{ -18 }{ 5 }x^-3 \] ?

Answer 6

so how is that a chain rule?

Answer 7

That what i dont get, it says apply chain rule. i dont know where.

Answer 8

I think you're missing an extra step, which does not come intuitively. It is something you need to be creative about

Answer 9

I dont even know.

Answer 10

what does chain rule look like? you need to make your original function look like something you can apply chain rule to

Answer 11

but it doesnt look like the chain rule would work there.

Answer 12

\(\huge g(x)=\frac{9}{5x^2} \rightarrow \frac{9}{\color{red}{\sqrt{25x^4}}}\)

Answer 13

does that look like something you can apply chain-rule to now?

Answer 14

how did you get the demoninator?
because i was taught that a Sqrt musnt be in the bottom?

Answer 15

it is not that SQUARE ROOT MUST NOT BE in the BOTTOM, it is just mathematically preferred for it in the numerator, I didn't really change anything. The value is still the same, I just looked for ways a chain-rule can be applied by first understanding what a chain-rule is and then make my problem looking similar to it without changing its value.

Answer 16

in calculus, you will see a lot of square roots in the denominator

Answer 17

Okay, understood.
Now you must use quotient rule correct? then apply chain rule? or is that wrong?

Answer 18

you're correct
in your quotient rule, \(\large g(x) = \frac{u}{v} \rightarrow \huge g'(x) = \frac{u'v-uv'}{v^2} \) the derivative of your denominator uses chain-rule.

Answer 19

How would you find the derivative of
\[\sqrt{25x^4}\]

Answer 20

use chain rule

Answer 21

LAUGHING OUT LOUD!

Answer 22

Lol, how? i dont understand

Answer 23

\(u=\sqrt{25 x^4} \rightarrow \huge (25x^4)^\frac{1}{2} \) chain-rule states that

we take the derivative of the outer function \( \huge\frac{1}{2}(25x^4)^{-\frac{1}{2}} \)

then multiply it by the derivative of the inner function \(\huge 4\times 25x^3 \)

giving us \( \color {blue}{\huge u' = \frac{1}{2}(25x^4)^{-\frac{1}{2}} \times 100x^3} \) now just simplify so it looks nice and clean

Answer 24

I thought you knew chain-rule already

Answer 25

I used u instead of v... just switch it around

Answer 26

In a simple problem i find it easy, but using it with quotient is harder for me

Answer 27

you can turn a quotient into a product

Answer 28

but i thought quotient need a demoninator?

Answer 29

ye
g(x) = u/v
u is the numerator
v is the denominator
so you have two separate functions that makes up the quotient, the u and the v

Answer 30

im lost.

Answer 31

I don't know why
have you learned quotient rule?

Answer 32

Recently yes.

Answer 33

so what is the problem and why are you lost?

Answer 34

do you remember what it is?

Answer 35

the way you did it above for the demoninator made is complex, and i dont understand what was happening/

Answer 36

your problem states to solve using chain rule
the way how it was originally written, chain rule cannot be used
so I made it look like something chain rule can be used without altering its value
it looks different now, but the value is still the same

it is like saying \(2 \times 2 = 2^2 \) they are written differently, but the value is the same
I just wrote the original problem into a format I can use chain rule

Answer 37

if you can't do fractional exponents and if algebra is not your strong suit, you will have a lot of trouble in calculus.

Answer 38

I've put a lot of time explaining already. It is time to move on.

Answer 39

I understand. Ugh a junior shouldnt be taking Ap-Calculus, it's very hard. Thanks for helping though.

Answer 40

you're a junior highschool, you should be taking calculus at that level like how other kids in other countries do.

Answer 41

you just need to sharpen up and keep practicing
go back to algebra or pre-calculus and make sure you master everything
I bet you that once you've achieved it, you won't have trouble in calculus

Answer 42

good luck
cheers

Answer 43

Thank You