Question

differentiate (sinx)^2 w.r.t. e^cosx .

Answer 1

what's w.r.t

Answer 2

"with respect to"

Answer 3

o

Answer 4

\[\large \frac{\partial}{\partial x} (\sin^2 (x))= \frac{\partial}{\partial x}(\sin^2(x)) \cdot \frac{\partial}{\partial x} (\sin (x))\]

Answer 5

So after that step...you can substitute.

Answer 6

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.

Answer 7

how could you equal both of them?? @Jhannybean ??

Answer 8

I'm not equaling anything? I'm expanding your equation.

Answer 9

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

Answer 10

Oops, made a mistake with my fraction. Ignore the second fraction.

Answer 11

Awesome @genius12

Answer 12

I forgot the 2 in the numerator for the final answer. Please add that.

@tanjeetsarkar96

Answer 13

olr8 got both of u .. @Jhannybean and @genius12 .

Answer 14

-2cos (x) / [e^(cos(x))]

Answer 15

yup, with the "2" which i forgot lol.

Answer 16

No problem,i forgot it's a parametric, and a function of a function.

Answer 17

@SithsAndGiggles , if you don't mind me asking, I'd like to know which grade you're in.

Answer 18

@Jhannybean Ya it's ok jhan, we can think of "x" as the parameter here instead of the conventional "t" =]