How do I derivate: x^(sin(x))?

Answer 1

Let y = x^(sin x)
ln y = ln x^(sin x)
ln y = (sin x)(ln x)
(1/y)(dy/dx) = (cos x)(ln x) + (sin x)/x
dy/dx = (cos x ln x + (sin x)/x) (x^(sin x))

Answer 2

general product rule is:

if you have a function that is the product of two functions of x, for instance, y(x) = f(x) * g(x), then the derivative is:

y'(x) = f(x)*g'(x) + g(x)*f '(x) equation 1

in this case, f(x) = x, and g(x) = sin(x).
so define the derivatives as f'(x) = 1, and g'(x) = cos(x)

Now plug these into equation 1.

y'(x) = x*cos(x) + 1*sin(x), which simplifies to
y'(x) = x*cos(x) + sin(x)

Answer 3

Think of it as y=x^(sin(x)). Then from this you can apply the ln rule:

lny=lnx^(sin(x))

Where this allows you to bring down the exponent function to get:

lny=(sin(x))lnx

Now, through implicit differentiation you can solve like this:

(1/y)(dy/dx)= cosx(lnx) + (1/x)(sinx)
dy/dx = y* [ cosx(lnx) + (1/x)(sinx) ]

Now you have this substitue in your y-value to get your final derivative functions!:

dy/dx = [x^(sin(x))]*[cosx(lnx) + (1/x)(sinx) ]

Answer 4

y = x^(sin x)

Take the log of both sides.

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

(1/y)(dy/dx) = (cos x)(ln x) + (sin x)/x

dy/dx = [(cos x)(ln x) + (sin x)/x] y

dy/dx = [(cos x)(ln x) + (sin x)/x] [x^(sin x)]