Question

Let f(x)=x^sinx For x>0
Find f'(x)

Answer 1

Let y = f(x)
y = x^sinx
ln y = sin x ln x
y'/y = (sin x)/x + cos x ln x
y' = x^sin x [(sin x)/x + cos x ln x]
f'(x) = y' = x^sin x [(sin x)/x + cos x ln x]

Answer 2

Im still confused, i dont know what properties to use.

Answer 3

Log properties. You can also change the method in the third step
ln y = sin x ln x
Put both sides to the power of e
y = e^(sin x ln x)
then differentiate from here if you prefer.

Answer 4

So i see why you ln them. But how did the y turn into y'?

Answer 5

Im having lots of trouble in indirect differentiation.

Answer 6

Implicit

Answer 7

Derivative of \[\large \frac{ d }{ dx } \ln f(x) = \frac{ f'(x) }{ f(x) }\]It's no different with y. \[\large \frac{ d }{ dx } \ln y = \frac{ y' }{ y}\]since the derivative of y (w.r.t. x) is y'

Answer 8

Alright thanks