fine i am puttin ma questions here

- anonymous

fine i am puttin ma questions here

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- schrodinger

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- anonymous

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

- anonymous

here i could not understand i got till \[d/dx(\log y) = d/dx(x^2 \log (\sin x))\] but am not able to proceed further

- anonymous

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

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## More answers

- anonymous

ok lets go slowly. your problem is that the variable is in the exponent.
you get it out of the exponent by taking the log

- anonymous

yups

- anonymous

so first step is to take the log

- anonymous

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

- anonymous

did that as u saw!!!

- anonymous

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

- anonymous

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

- anonymous

so far so good?

- anonymous

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

- anonymous

see above !!!!!

- anonymous

satellite: I suggest you to start using a basic chain rule example as he stumbles only upon the chain part! good luck

- anonymous

ok next step is to take the derivative.
\[\frac{d}{dx}x^2ln(sin(x))\]
and for this you need both the product rule and the chain rule.

- anonymous

dude i did this already!!!!

- anonymous

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

- anonymous

as it is both a product,
\[x^2\times ln(sin(x))\]
and
\[ln(sin(x))\] is a composite function. is the the log OF the sine of x

- anonymous

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

- anonymous

in english, the derivative of a composite function is the derivative of the outside function evaluated at the inside function, times the derivative of the inside function.

- anonymous

i got confused!!!!!!

- anonymous

composite means \[f(g(x))\]
so if \[f(x)=ln(x)\]
and \[g(x) = sin(x)\]
then \[f(g(x))=ln(g(x))=ln(sin(x))\]

- anonymous

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

- anonymous

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

- anonymous

this is a composition of functions. another example.
\[f(x)=\sqrt{x}\]
\[g(x)=1-x^2\]
\[f(g(x))=f(1-x^2)=\sqrt{1-x^2}\]

- anonymous

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

- anonymous

I am saying the same for th LHS ie log y

- anonymous

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

- anonymous

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

- anonymous

i am going slow and still wanting to take the derivative of
\[x^2ln(sin(x))\] correctly. we will deal with the consequence of taking the log later.

- anonymous

yes here is where i am stuck

- anonymous

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

- anonymous

right

- anonymous

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

- anonymous

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

- anonymous

product rule says dat 1st take the 1st constant and diff the other and add the vis - a - vis to it!!!!!!

- anonymous

nope sorry. again lets to slow

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yups i know f'(x)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ahh that is the "chain rule"

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

right

- anonymous

times the derivative of sine(x)

- anonymous

which is...

- anonymous

a rule?????????

- anonymous

do you know the derivative of sin(x)?

- anonymous

nope

- anonymous

guess.

- anonymous

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

- anonymous

is it cos(x)

- anonymous

yes!

- anonymous

so i repeat
but the derivative of log of "something" is one over something times the derivative of " something"

- anonymous

so we multiply the cos(x)

- anonymous

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

- anonymous

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

- anonymous

whew got it.

- anonymous

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

- anonymous

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

- anonymous

now we put this all back in the product rule
(FG)'=F'G+G'F
\[F+x^2\]
\[F'=2x\]
\[G=ln(sin(x))\]
\[G'=\frac{cos(x)}{sin(x)}\]

- anonymous

i meant \[F=x^2\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes!!!!!!!

- anonymous

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

- anonymous

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

- anonymous

we did not find the derivative of the original thing, we found the derivative of the log of the original thing.

- anonymous

so wat next?

- anonymous

ok i will say quickly what we do next because it is easy. then i will give an explanation. next we simply get our answer by multiplying what we got above by the original function. done.

- anonymous

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

- anonymous

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

- anonymous

gotta go!!!!!!!

- anonymous

the first part is the original function you were to differentiate. the second part was the derivative of the log of that
good luck

- anonymous

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

- anonymous

ok final step. don't forget that the first step was taking the log. so we took the derivative of
\[ln(f(x))\] not \[f(x)\]
by the chain rule the derivative of
\[ln(f(x))\]
is
\[\frac{1}{f(x)}\times f'(x)=\frac{f'(x)}{f(x)}\]
that is
\[\frac{d}{dx}ln(f(x))=\frac{f'(x)}{f(x)}\]
solving this equation for \[f'(x)\]
gives \[f'(x)=f(x)\times \frac{d}{dx}ln(f(x))\]
in english you can find the derivative by first taking the log of your function, then differentiating the log of your function, and then multiplying by the original function.
hope this helps

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