Question

Differentiate the following:

Answer 1

\[\cos(x ^{2}+2y)=xe ^{y ^{2}}\]

Answer 2

use chain rule

Answer 3

For both parties?

Answer 4

yeah

Answer 5

in right hand side product rule

Answer 6

for both "parties" .. lol!

Answer 7

\[\cos(2x+2) + \sin(x ^{2}+2y) = xe ^{y ^{2}}(2y)+e ^{y ^{2}}\]
right?

Answer 8

\[\frac{ d }{ dx}\cos(x^2+2y)=[-\sin(x^2+2y)*\frac{ d }{dx }(x^2+2y)]\]

Answer 9

so
\[LHS=[-\sin(x^2+2y)*(2x+2\frac{ dy }{ dx })]\]
ok with this

Answer 10

But I think it will be a plus in product rule not Multiplication

\[\frac{ d }{ dx}\cos(x^2+2y)=[-\sin(x^2+2y)+\frac{ d }{dx }(x^2+2y)]\]

Answer 11

no the chain rule states that
\[\large \large \frac{ d }{ dx }[f[g(x)]=f ^{'}[g(x)]*g^{'}(x)\]

Answer 12

yes for product rule there will be plus but for chain rule there will be multiplication

Answer 13

aha .. sorry about that

so we have finished LHS ?

Answer 14

ok LHS is finished
now for RHS you made a bit mistake it will be
\[\frac{ d }{ dx }[x*e^{y^{2}}]=x*\frac{ d }{ dx }[e^{y^{2}}]+e^{y^{2}}*\frac{ d }{ dx }(x)\]
ok??

Answer 15

now \[Rhs=x*[e^{y^{2}}*2y*\frac{ dy }{ dx}]+e^{y^{2}}\]
ok with this??

Answer 16

u forgot to write dy/dx ,because we are differentiating with respect to x

Answer 17

product rule right?

Answer 18

yeah
and when ure differentiating e^(y^2) u have to use chain rule again
\[\frac{ d }{ dx }[e^{y^{2}}]=e^{y^{2}}*\frac{ d }{ dx }(y^2)=e^{y^{2}}*2y \frac{ dy }{ dx }\]

Answer 19

yeah I know that ..

and what's next ?

Answer 20

now both sides are completed
now take the terms with (dy/dx) in one side and other terms in one side then take (dy/dx) common and u'll be finished

Answer 21

ok wait me I will do and I will show you ..

Answer 22

\[2x+2y \prime . 2y.y \prime = \sin(x ^{2}+2y).x.e ^{y ^{2}}+e ^{y ^{2}}\]

so will come like that?

Answer 23

ur doing some thing wrong check again

Answer 24

\[2y \prime . 2y.y \prime = \sin(x ^{2}+2y).x.e ^{y ^{2}}+e ^{y ^{2}}-2x\]

now it's true?