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satellite73Best ResponseYou've already chosen the best response.1
\[f(x)=\frac{\sin^2(x^2+5)}{\sqrt{x^2+1}}\]? like that? will be an algebra nightmare your math teacher must hate you
 one year ago

nickymardenBest ResponseYou've already chosen the best response.0
haha, that's right. I was supposed t do it in class, but i got to the middle and got fed up. Can you help me? please..
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
\[(\frac{f}{g})'=\frac{gf'f'g}{g^2}\] with \[f(x)=\sin^2(x^2+5),g(x)=\sqrt{x^2+1}\] we need both \(f'\) and \(g'\) and both require the chain rule
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
so first we note that \(f\) is actually a composition of three functions, \(x^2+5\) , sine and the squaring function so good \[f'(x)=2\sin(x^2+5)\times \cos(x^2+5)\times 2x=4x\sin(x^2+5)\cos(x^2+5)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and also by the chain rule \[g'(x)=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}\] so far so good?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now we have to put this mess together for \[(\frac{f}{g})'=\frac{gf'f'g}{g^2}\] the denominator is easy enough, it is just \(x^2+1\) but the numerator is going to be a colossal mess i guess we can just write it down
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
numerator is \[\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
and so your whole mess is \[\frac{\sqrt{x^2+1}\times 4x\sin(x^2+5)\cos(x^2+5)\sin^2(x^2+5)\times \frac{x}{\sqrt{x^2+1}}}{x^2+1}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now you can get rid of the compound fraction by multiplying top and bottom by \(\sqrt{x^2+1}\) to get \[\frac{(x^2+1)\times 4x\sin(x^2+5)\cos(x^2+5)\sin^2(x^2+5)\times x}{(x^2+1)\sqrt{x^2+1}}\] and maybe you can even clean this mess up with some more algebra by multiplying out in the numerator or something, but my algebra is not that good
 one year ago

nickymardenBest ResponseYou've already chosen the best response.0
omg, if i could give you a million medals i would.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
there is also an annoying trig identity you can use since \[2\sin(x)\cos(x)=\sin(2x)\] so if you want to be really clever you can replace \[4x\sin(x^2+5)\cos(x^2+5)\] by \[2x\sin(2x^2+10)\] but i wouldn't
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
one is more than enough yw
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh also you can write the denominator as \((x^2+1)^{\frac{3}{2}}\) but that strikes me as unnecessary also
 one year ago

nickymardenBest ResponseYou've already chosen the best response.0
but he wants it like that lol thank you :)
 one year ago
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