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OpenStudy (anonymous):

calculus

OpenStudy (anonymous):

If y = u^3 and x=5u^2 . The value of dy/dx at u = 10 is

OpenStudy (anonymous):

@mathslover

OpenStudy (anonymous):

@iambatman

OpenStudy (vishweshshrimali5):

FInd out \(\cfrac{dy}{du}\) and \(\cfrac{dx}{du}\) Divide them, you will get \(\cfrac{dy}{dx}\)/

OpenStudy (anonymous):

i have encountered such problem for the first time

OpenStudy (anonymous):

dy/dx = 3x^2

OpenStudy (vishweshshrimali5):

Plug in the value of u and you will get your answer. Actually such type of differentiation is called parametric differentiation.

OpenStudy (vishweshshrimali5):

You must obtain the value of dy/dx in terms of u.

OpenStudy (anonymous):

dx/du = 10u

OpenStudy (anonymous):

3x^2/10u 3x^2/20

OpenStudy (vishweshshrimali5):

Yeah

OpenStudy (vishweshshrimali5):

Now what is dy/du

OpenStudy (marissalovescats):

I feel like I am now ahead in the Calculus world by learning. I didn't learn this in AP Calc AB.. But maybe next year in AP Calc BC but now have a clue about so woo head start oh yeah.

OpenStudy (marissalovescats):

Words are missing from my sentence, but you get the idea. No judgement I just woke up.

OpenStudy (anonymous):

Never encountered such problem before , thanks @vishweshshrimali5 and @iambatman

OpenStudy (anonymous):

answer is 3 actually

OpenStudy (anonymous):

:(

OpenStudy (vishweshshrimali5):

Yeah that's correct.

OpenStudy (anonymous):

how?

OpenStudy (vishweshshrimali5):

@No.name what is the value of dy/du ?

OpenStudy (anonymous):

3x^2/20

OpenStudy (vishweshshrimali5):

Please tell me the value in terms of u only

OpenStudy (anonymous):

3x^2/10u

OpenStudy (vishweshshrimali5):

See, its nothing different than dy/dx. Just ignore that you have x. You only have y in terms of u. I am asking you to differentiate y with respect to u.

OpenStudy (vishweshshrimali5):

I am writing the question again for you: \(\color{blue}{\text{Originally Posted by}}\) @No.name If y = u^3 The value of dy/du at u = 10 is \(\color{blue}{\text{End of Quote}}\)

OpenStudy (vishweshshrimali5):

@No.name can you find out the answer of this question ?

myininaya (myininaya):

\[\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}\] finding dy/du is pretty straight forward. you can find du/dx by first finding dx/du.

OpenStudy (vishweshshrimali5):

@No.name , can you find out dy/du or not ?

OpenStudy (anonymous):

3u^2 ?

OpenStudy (vishweshshrimali5):

Yes Now try finding out dx/du.

OpenStudy (anonymous):

10u

OpenStudy (vishweshshrimali5):

Great

OpenStudy (anonymous):

Sorry my net is not functioning well , so taking time to reply

OpenStudy (vishweshshrimali5):

Now, \[\large{\cfrac{dy}{dx} = \cfrac{\cfrac{dy}{du}}{\cfrac{dx}{du}}}\]

OpenStudy (vishweshshrimali5):

Now can you find out dy/dx ?

OpenStudy (anonymous):

3u^2/10u 3u/10

OpenStudy (vishweshshrimali5):

Good put u = 10

OpenStudy (anonymous):

3

OpenStudy (vishweshshrimali5):

Great there you have it !!

OpenStudy (anonymous):

OK I GET IT NOW

OpenStudy (anonymous):

THANKS :)

OpenStudy (anonymous):

Good job @No.name and nice teaching @vishweshshrimali5 :)

OpenStudy (vishweshshrimali5):

No problem. Enjoy your time with calculus it really is very interesting @No.name @iambatman that is one of the very few things I am good at :P

OpenStudy (anonymous):

One more thing i wanted to ask is what is the derivative of e^-x

OpenStudy (anonymous):

and how do you find it

OpenStudy (vishweshshrimali5):

-e^{-x}

OpenStudy (anonymous):

e^(x squared) ?

OpenStudy (vishweshshrimali5):

First find out the derivative of power and then calculate the derivative of the base (here e).

OpenStudy (vishweshshrimali5):

That is use chain rule.

OpenStudy (anonymous):

e^(x squared)2x

OpenStudy (vishweshshrimali5):

The same use product rule and then chain rule for each.

OpenStudy (vishweshshrimali5):

Calculus is all about practicing. You should be able to solve calculus problems even while sleeping,

OpenStudy (vishweshshrimali5):

Bye friends :)

OpenStudy (anonymous):

Good bye

OpenStudy (marissalovescats):

Wait isn't the derivative of e^-x -e^-x?

OpenStudy (anonymous):

Yes it is

OpenStudy (marissalovescats):

Oh okay thought so lol

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