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1st of all in the question \\[y = sinx ^{x ^{2}}\]

The chain rule states that,
\[d/dx(f(y)) = f'(y) dy/dx\]

yups

so first step is to take the log

now d/dx(logy) = 1/y dy/dx? Did you get this?

did that as u saw!!!

no amogh i did not get this !!!!!

\[ln(sin(x))^{x^2}=x^2ln(sin(x))\] by the properties of logs

so far so good?

that's wat i did till here i got !!!!!!

see above !!!!!

dude i did this already!!!!

so now i apply chain rule????????

wat do u mean by a composite function??????

i got confused!!!!!!

wat happened to the outer x^2????

when i diff (log y )i get i/y

oh no.
the derivative of
\[ln(x)=\frac{1}{x}\]

I am saying the same for th LHS ie log y

but the derivative of \[ln(g(x))=\frac{1}{g(x)}\times g'(x)\]

so the derivative of \[ln(sin(x))=\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]

yes here is where i am stuck

x^2 becomes 2x right???????

right

you have a product here. do you know the product rule?

yay so we get in the end 2x.(ln(sin(x)))

nope sorry. again lets to slow

CHEATING!!!!!!! ( cranky kid's voice)!!!!!!!!

the derivative of \[f(x)g(x)\] is \[f'(x)g(x)+g'(x)g(x)\]

well dat's wat i said in words just to avoid the typing!!!!!!!

put \[f(x)=x^2\]
\[f'(x)=2x\]

yups i know f'(x)

\[g(x)=ln(sin(x))\]

\[g'(x)=\frac{cos(x)}{sin(x)}\]

now dis is de place where i get stuck up y do we get dis????????????

ahh that is the "chain rule"

i say it in english. the derivative of log x is one over x

how is this the chain rule i mean i just read it for algebra not for log or sin

but the derivative of log of "something" is one over something times the derivative of "
something"

so by dat logic ln(sin(x) ) when derivatived we shud get 1/sin(x)

right

times the derivative of sine(x)

which is...

a rule?????????

do you know the derivative of sin(x)?

nope

guess.

i know dat u r a gud teacher and u r real patient thanx for dat!!!!!!

is it cos(x)

yes!

so we multiply the cos(x)

ok ca we quickly repeat!!!!!!!!!!!!!

whew got it.

so we get 1/y = 2x. cos(x)/sin(x) true???

srsly whew my brain is alrdy stewd!!!!!

i meant \[F=x^2\]

yes go on!!!!! i am trying to get a full pic

to get ....
the derivative of
\[x^2ln(sin(x))=2xln(sin(x))+x^2 \frac{cos(x)}{sin(x)}\]

that is the product rule, although we used the chain rule to find the derivative of \[ln(sin(x))\]

yes!!!!!!!

whew. of course we are not done! believe it or not. because we took the log as the first step

i cud not pick that up!!!!! that ln(sin(x) ) was by a rule??????????

so wat next?

CONFUSION HERE IN WAT U JUST SAID (short circuited)!!!!!!!!!!!!!!!!!!!!!!!!

answer is...
\[(sin(x))^{x^2}\times (2xln(sin(x))+\frac{x^2cos(x)}{sin(x)})\]

gotta go!!!!!!!

i can 't understand the final step on how to obtain the answer