Question

find dy/dx of sin(x)^x

Answer 1

(xcos(x) - sin(x))/x^2 using quotient rule

Answer 2

y = (sin(x))^x


ln(y) = ln((sin(x))^x)


ln(y) = x*ln(sin(x))


d/dx[ln(y)] = d/dx[x*ln(sin(x))]


y'/y = ln(sin(x))+(x*cos(x))/sin(x)


y'/y = ln(sin(x))+x*cot(x)


y' = y(ln(sin(x))+x*cot(x))


y' = (sin(x))^x(ln(sin(x))+x*cot(x))

Answer 3

how did you know you had to do it that way?

Answer 4

forget my ans -- I solved a similar one.

Answer 5

it's in the form f(x)^(g(x)), so you have to use a log

Answer 6

ohh alright lol ty

Answer 7

you can also write
\[\sin(x)^x=e^{x\ln(\sin(x))}\] and use the chain rule but the work is identical

Answer 8

haha alright do you know why he did

Answer 9

ln(sinx)+ xcos/sin?

Answer 10

i dont understand why he has the + sign in there

Answer 11

yes it is the product rule+ chain rule

Answer 12

why is that the first part ?

Answer 13

yes I used the product rule which is

d/dx[f(x)*g(x)] = f'(x)g(x) + f(x)g'(x)


then the chain rule


d/dx[f(g(x))] = g'(x)*f'(g(x))

Answer 14

you have
\[x\ln(\sin(x))\] which is a product. so you have to take the derivative of x (which is 1) and multiply it by
\[\ln(\sin(x))\] and then take x and multiply it by the derivative of
\[\ln(\sin(x))\] which is
\[\frac{1}{\sin(x)}\times \cos(x)=\frac{\cos(x)}{\sin(x)}=\cot(x)\]

Answer 15

hopefully that's a bit clearer...

Answer 16

ohhh ya that helps alot actually
ty guys