anonymous
  • anonymous
fine i am puttin ma questions here
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
1st of all in the question \\[y = sinx ^{x ^{2}}\]
anonymous
  • 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
  • anonymous
The chain rule states that, \[d/dx(f(y)) = f'(y) dy/dx\]

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anonymous
  • 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
  • anonymous
yups
anonymous
  • anonymous
so first step is to take the log
anonymous
  • anonymous
now d/dx(logy) = 1/y dy/dx? Did you get this?
anonymous
  • anonymous
did that as u saw!!!
anonymous
  • anonymous
no amogh i did not get this !!!!!
anonymous
  • anonymous
\[ln(sin(x))^{x^2}=x^2ln(sin(x))\] by the properties of logs
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
that's wat i did till here i got !!!!!!
anonymous
  • anonymous
see above !!!!!
anonymous
  • anonymous
satellite: I suggest you to start using a basic chain rule example as he stumbles only upon the chain part! good luck
anonymous
  • 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
  • anonymous
dude i did this already!!!!
anonymous
  • anonymous
so now i apply chain rule????????
anonymous
  • 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
  • anonymous
wat do u mean by a composite function??????
anonymous
  • 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
  • anonymous
i got confused!!!!!!
anonymous
  • 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
  • anonymous
wat happened to the outer x^2????
anonymous
  • anonymous
when i diff (log y )i get i/y
anonymous
  • 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
  • anonymous
oh no. the derivative of \[ln(x)=\frac{1}{x}\]
anonymous
  • anonymous
I am saying the same for th LHS ie log y
anonymous
  • anonymous
but the derivative of \[ln(g(x))=\frac{1}{g(x)}\times g'(x)\]
anonymous
  • anonymous
so the derivative of \[ln(sin(x))=\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]
anonymous
  • 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
  • anonymous
yes here is where i am stuck
anonymous
  • anonymous
x^2 becomes 2x right???????
anonymous
  • anonymous
right
anonymous
  • anonymous
you have a product here. do you know the product rule?
anonymous
  • anonymous
yay so we get in the end 2x.(ln(sin(x)))
anonymous
  • anonymous
product rule says dat 1st take the 1st constant and diff the other and add the vis - a - vis to it!!!!!!
anonymous
  • anonymous
nope sorry. again lets to slow
anonymous
  • anonymous
CHEATING!!!!!!! ( cranky kid's voice)!!!!!!!!
anonymous
  • anonymous
the derivative of \[f(x)g(x)\] is \[f'(x)g(x)+g'(x)g(x)\]
anonymous
  • anonymous
well dat's wat i said in words just to avoid the typing!!!!!!!
anonymous
  • anonymous
put \[f(x)=x^2\] \[f'(x)=2x\]
anonymous
  • anonymous
yups i know f'(x)
anonymous
  • anonymous
\[g(x)=ln(sin(x))\]
anonymous
  • anonymous
\[g'(x)=\frac{cos(x)}{sin(x)}\]
anonymous
  • anonymous
now dis is de place where i get stuck up y do we get dis????????????
anonymous
  • anonymous
ahh that is the "chain rule"
anonymous
  • anonymous
i say it in english. the derivative of log x is one over x
anonymous
  • anonymous
how is this the chain rule i mean i just read it for algebra not for log or sin
anonymous
  • anonymous
but the derivative of log of "something" is one over something times the derivative of " something"
anonymous
  • anonymous
so by dat logic ln(sin(x) ) when derivatived we shud get 1/sin(x)
anonymous
  • anonymous
right
anonymous
  • anonymous
times the derivative of sine(x)
anonymous
  • anonymous
which is...
anonymous
  • anonymous
a rule?????????
anonymous
  • anonymous
do you know the derivative of sin(x)?
anonymous
  • anonymous
nope
anonymous
  • anonymous
guess.
anonymous
  • anonymous
i know dat u r a gud teacher and u r real patient thanx for dat!!!!!!
anonymous
  • anonymous
is it cos(x)
anonymous
  • anonymous
yes!
anonymous
  • anonymous
so i repeat but the derivative of log of "something" is one over something times the derivative of " something"
anonymous
  • anonymous
so we multiply the cos(x)
anonymous
  • anonymous
ok ca we quickly repeat!!!!!!!!!!!!!
anonymous
  • anonymous
so we get the derivative of \[ln(sin(x))\] is \[\frac{1}{sin(x)}\times cos(x)=\frac{cos(x)}{sin(x)}\]\]
anonymous
  • anonymous
whew got it.
anonymous
  • anonymous
so we get 1/y = 2x. cos(x)/sin(x) true???
anonymous
  • anonymous
srsly whew my brain is alrdy stewd!!!!!
anonymous
  • 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
  • anonymous
i meant \[F=x^2\]
anonymous
  • anonymous
yes go on!!!!! i am trying to get a full pic
anonymous
  • anonymous
to get .... the derivative of \[x^2ln(sin(x))=2xln(sin(x))+x^2 \frac{cos(x)}{sin(x)}\]
anonymous
  • anonymous
that is the product rule, although we used the chain rule to find the derivative of \[ln(sin(x))\]
anonymous
  • anonymous
yes!!!!!!!
anonymous
  • anonymous
whew. of course we are not done! believe it or not. because we took the log as the first step
anonymous
  • anonymous
i cud not pick that up!!!!! that ln(sin(x) ) was by a rule??????????
anonymous
  • anonymous
we did not find the derivative of the original thing, we found the derivative of the log of the original thing.
anonymous
  • anonymous
so wat next?
anonymous
  • 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
  • anonymous
CONFUSION HERE IN WAT U JUST SAID (short circuited)!!!!!!!!!!!!!!!!!!!!!!!!
anonymous
  • anonymous
answer is... \[(sin(x))^{x^2}\times (2xln(sin(x))+\frac{x^2cos(x)}{sin(x)})\]
anonymous
  • anonymous
gotta go!!!!!!!
anonymous
  • 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
  • anonymous
i can 't understand the final step on how to obtain the answer
anonymous
  • 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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