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fine i am puttin ma questions here

Mathematics
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1st of all in the question \\[y = sinx ^{x ^{2}}\]
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
The chain rule states that, \[d/dx(f(y)) = f'(y) dy/dx\]

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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
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 !!!!!
satellite: I suggest you to start using a basic chain rule example as he stumbles only upon the chain part! good luck
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.
dude i did this already!!!!
so now i apply chain rule????????
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
wat do u mean by a composite function??????
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.
i got confused!!!!!!
composite means \[f(g(x))\] so if \[f(x)=ln(x)\] and \[g(x) = sin(x)\] then \[f(g(x))=ln(g(x))=ln(sin(x))\]
wat happened to the outer x^2????
when i diff (log y )i get i/y
this is a composition of functions. another example. \[f(x)=\sqrt{x}\] \[g(x)=1-x^2\] \[f(g(x))=f(1-x^2)=\sqrt{1-x^2}\]
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 \[ln(sin(x))=\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]
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.
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)))
product rule says dat 1st take the 1st constant and diff the other and add the vis - a - vis to it!!!!!!
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 i repeat but the derivative of log of "something" is one over something times the derivative of " something"
so we multiply the cos(x)
ok ca we quickly repeat!!!!!!!!!!!!!
so we get the derivative of \[ln(sin(x))\] is \[\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]\]
whew got it.
so we get 1/y = 2x. cos(x)/sin(x) true???
srsly whew my brain is alrdy stewd!!!!!
now we put this all back in the product rule (FG)'=F'G+G'F \[F+x^2\] \[F'=2x\] \[G=ln(sin(x))\] \[G'=\frac{cos(x)}{sin(x)}\]
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??????????
we did not find the derivative of the original thing, we found the derivative of the log of the original thing.
so wat next?
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.
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!!!!!!!
the first part is the original function you were to differentiate. the second part was the derivative of the log of that good luck
i can 't understand the final step on how to obtain the answer
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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