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Mathematics 14 Online
OpenStudy (anonymous):

Differentiation

OpenStudy (anonymous):

y = (4 - 3x)^9

OpenStudy (fibonaccichick666):

slowpoke ;P

OpenStudy (fibonaccichick666):

ok so remember how the chain rule works?

OpenStudy (fibonaccichick666):

what is your u=? du=?

OpenStudy (fibonaccichick666):

then compute again

OpenStudy (anonymous):

can u solve this problem for me please in dy/dx form

OpenStudy (fibonaccichick666):

no.. You know I don't do that

OpenStudy (fibonaccichick666):

remember dy/dx=f'(x)

OpenStudy (fibonaccichick666):

they are equivalent

OpenStudy (anonymous):

I wan a direct answer actually by that i can solve future problems { I know the code of conduct} BUT PLEASE I BEG YOU

OpenStudy (anonymous):

I will understand it you don't worry

OpenStudy (fibonaccichick666):

noooo nevaaaaa

OpenStudy (fibonaccichick666):

so u=? du=?

ganeshie8 (ganeshie8):

how come u have differentiation in 10th grade ?

OpenStudy (anonymous):

I am doing IIT do they started teaching it in another class

ganeshie8 (ganeshie8):

\(y = (4 - 3x)^9\) say \(u = 4-3x\)

OpenStudy (anonymous):

Because this level calculus problems is too kiddish for iit

ganeshie8 (ganeshie8):

\(y = (u)^9\)

OpenStudy (anonymous):

ok

ganeshie8 (ganeshie8):

knw how to differentiate this funciton with respect to \(u\) ?

OpenStudy (anonymous):

9u^8

ganeshie8 (ganeshie8):

yes ! but u dont want just dy/du. u want dy/dx

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

use chain rule : \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)

ganeshie8 (ganeshie8):

\(y = (u)^9\) , \(u = 4-3x\) \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\) \(~~~~~= 9u^8 \times \dfrac{du}{dx}\)

ganeshie8 (ganeshie8):

find out du/dx and plugin

OpenStudy (anonymous):

I think i got it wait

ganeshie8 (ganeshie8):

yes, these are easy to do on one's own. explaining is bit tricky..

OpenStudy (anonymous):

How to differentiate u d/dx of 4 would be 0

ganeshie8 (ganeshie8):

yes

ganeshie8 (ganeshie8):

\(u = 4-3x\) \(\dfrac{du}{dx} = 0 - 3 \)

OpenStudy (anonymous):

-3

OpenStudy (anonymous):

so, 9x*8 * -3 -27^8 ?

OpenStudy (anonymous):

9x^8 i meant

ganeshie8 (ganeshie8):

\(y = (u)^9 \) , \(u = 4-3x\) \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\) \(~~~~~= 9u^8 \times (-3)\)

ganeshie8 (ganeshie8):

plug back the value of \(u\) ^

OpenStudy (anonymous):

27(4 - 3x )^8

ganeshie8 (ganeshie8):

lol who ate the negative sign

OpenStudy (anonymous):

yeah typo lol

ganeshie8 (ganeshie8):

\(y = (u)^9 \) , \(u = 4-3x\) \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\) \(~~~~~= 9u^8 \times (-3)\) \(~~~~~= 9(4-3x)^8 \times (-3)\) \(~~~~~= -27(4-3x)^8\)

ganeshie8 (ganeshie8):

^^thats how u do it step by step. but once u get practice, u can do it in single line

ganeshie8 (ganeshie8):

\(y = (4 - 3x)^9 \) \(\dfrac{dy}{dx} = 9(4-3x)^8 (4-3x)' = 9(4-3x)^8(-3)\)

OpenStudy (anonymous):

Thank you once again

ganeshie8 (ganeshie8):

np :)

OpenStudy (anonymous):

(thank you )^infinity

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