Mathematics
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OpenStudy (anonymous):
Differentiation
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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
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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
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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
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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 \(u\) ?
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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):
\(y = (u)^9\) , \(u = 4-3x\)
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)
\(~~~~~= 9u^8 \times \dfrac{du}{dx}\)
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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
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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):
\(y = (u)^9 \) , \(u = 4-3x\)
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)
\(~~~~~= 9u^8 \times (-3)\)
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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):
\(y = (u)^9 \) , \(u = 4-3x\)
\(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\)
\(~~~~~= 9u^8 \times (-3)\)
\(~~~~~= 9(4-3x)^8 \times (-3)\)
\(~~~~~= -27(4-3x)^8\)
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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