anonymous
  • anonymous
differentiate (sinx)^2 w.r.t. e^cosx .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what's w.r.t
anonymous
  • anonymous
"with respect to"
anonymous
  • anonymous
o

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Jhannybean
  • Jhannybean
\[\large \frac{\partial}{\partial x} (\sin^2 (x))= \frac{\partial}{\partial x}(\sin^2(x)) \cdot \frac{\partial}{\partial x} (\sin (x))\]
Jhannybean
  • Jhannybean
So after that step...you can substitute.
anonymous
  • anonymous
Chain rule? \[\frac{d~f(x)}{d~g(x)}=\frac{\frac{d~f(x)}{dx}}{\frac{dx}{d~g(x)}}=\frac{df}{dx}\cdot\frac{dx}{dg}\] Where \(f(x)=\sin x\), and \(g(x)=e^{\cos x}\). I'm not totally sure about this, though.
anonymous
  • anonymous
how could you equal both of them?? @Jhannybean ??
Jhannybean
  • Jhannybean
I'm not equaling anything? I'm expanding your equation.
anonymous
  • anonymous
So if we a function letter to each we get:\[\bf y=\sin^2(x)\]and\[\bf z=e^{\cos(x)}\]We are finding \(\bf \frac{dy}{dz}\), and to do that, we will need to use Chain Rule the way it's used for differentiating parametric functions. If y is considered a function of z then:\[\bf \frac{ dy }{ dz }=\frac{ \frac{ dy }{ dx } }{ \frac{ dz }{ dx } }\]To compute the derivative, we plug in the derivatives for y and z:\[\bf \frac{ dy }{ dz }=\frac{ 2\cancel{\sin(x)}\cos(x) }{ -e^{\cos(x)}\cancel{\sin(x)} }=-\frac{ \cos(x) }{ e^{\cos(x)} }\] @tanjeetsarkar96
anonymous
  • anonymous
Oops, made a mistake with my fraction. Ignore the second fraction.
Jhannybean
  • Jhannybean
Awesome @genius12
anonymous
  • anonymous
I forgot the 2 in the numerator for the final answer. Please add that. @tanjeetsarkar96
anonymous
  • anonymous
olr8 got both of u .. @Jhannybean and @genius12 .
Jhannybean
  • Jhannybean
-2cos (x) / [e^(cos(x))]
anonymous
  • anonymous
yup, with the "2" which i forgot lol.
Jhannybean
  • Jhannybean
No problem,i forgot it's a parametric, and a function of a function.
anonymous
  • anonymous
@SithsAndGiggles , if you don't mind me asking, I'd like to know which grade you're in.
anonymous
  • anonymous
@Jhannybean Ya it's ok jhan, we can think of "x" as the parameter here instead of the conventional "t" =]

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