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Mathematics 20 Online
OpenStudy (anonymous):

differentiate

hartnn (hartnn):

do you know product rule and chain rule ?

OpenStudy (anonymous):

i know product rule..chain rule i understand a little

hartnn (hartnn):

ok, so give it a shot, use product rule to differentiate : x^4y^5 what u get ?

OpenStudy (anonymous):

4x^3*y^5+x^4*5y^4

hartnn (hartnn):

when you differentiate any function of 'y' , with respect to 'x' , like say f(y), you use chain rule and get f'(y) dy/dx similarly when you differentiated y^5 here, you would get 5y^4 dy/dx got this ?

OpenStudy (anonymous):

yeah so did i do the above correctly?

hartnn (hartnn):

just multiply the last term by dy/dx everything else is correct

OpenStudy (anonymous):

why the last term? 5y^4?

hartnn (hartnn):

\(4x^3y^5+x^4 5y^4D\) that should go into your result box :)

hartnn (hartnn):

yes, because you differentiated the function of y aka y^5 only in 2nd term, so dy/dx will be multiplied only in 2nd term

hartnn (hartnn):

\(4x^3y^5+5x^4 y^4D\)

OpenStudy (anonymous):

okay i kind of understand it

hartnn (hartnn):

if you practice a few more problems you will get it clearly :) as its not difficult at all.

OpenStudy (anonymous):

oh okay but it says to respect to x...why are we focusing on y?

hartnn (hartnn):

since we r diff w.r.t x, thats why we need chain rule for differentiating a function of 'y' ! else if we differentiate a function of 'x' say x^9, we won't need chain rule, it'll be just 9x^8 but for y^9, differentiating w.r.t x would give 9y^8 dy/dx

OpenStudy (anonymous):

okay hmm im not clearly understanding it but i will eventually! thanks again!! :)

hartnn (hartnn):

similarly if we differentiate x^9 w.r.t y , we would get 9x^8 dx/dy ok, welcome ^_^

OpenStudy (anonymous):

@hartnn im having a problem with another example

hartnn (hartnn):

sure, ask :)

OpenStudy (anonymous):

(x^2+y^2)^7

OpenStudy (anonymous):

i did what you told me but its not working

hartnn (hartnn):

whats your answer ?

OpenStudy (anonymous):

i got 7(x^2+y^2)^6*2x*Dy^2+2y*x^2

OpenStudy (anonymous):

is the answer close? or very far lol

hartnn (hartnn):

7(x^2+y^2)^6 part is correct ....

OpenStudy (anonymous):

the second part is g'(x)

hartnn (hartnn):

then d/dx (x^2+y^2) = (2x +2y D)

hartnn (hartnn):

thats it!

OpenStudy (anonymous):

what happend to the other part?

hartnn (hartnn):

i mean 7(x^2+y^2)^6 (2x +2y D)

OpenStudy (anonymous):

oh okay i think i differtiated that part wrong

hartnn (hartnn):

yeah, diff of x^2 is just 2x and y^2 is just 2ydy/dx

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