What is the derivative of #x^sin(x)#?

#y = x^sinx#

Take the natural logarithm of both sides.

#lny = ln(x^sinx)#

Use laws of logarithms to simplify.

#lny = sinxlnx#

Use the product rule and implicit differentiation to differentiate.

#1/y(dy/dx) = cosx(lnx) + 1/x(sinx)#

#1/y(dy/dx) = cosxlnx + sinx/x#

#dy/dx = (cosxlnx + sinx/x)/(1/y)#

#dy/dx = x^sinx(cosxlnx+sinx/x)#

Hopefully this helps!

When we have a function of #x# like #y=x^sin x#, where a single term contains #x# in both its base and its power, perhaps the easiest way to find the function's derivative is to first take the (natural) logarithm of both sides:

#ln y = ln (x^(sin x))#
#color(white)(ln y)=sin x * ln x#

This places all the #x#'s on the same "level". Then, take the derivative of both sides with respect to #x#:

#=>d/dx (ln y)=d/dx (sin x * ln x)#

Remembering that #y# is a function of #x#, we get

#=> 1/y*dy/dx=cos x * ln x + sin x (1/x)#
#=> color(white)"XXi"dy/dx=y[cos x * ln x + (sin x) /x]#

Since we began with #y=x^(sin x)#, we substitute this back in for #y# to get

#=> color(white)"XXi"dy/dx=x^(sin x)[cos x * ln x + (sin x)/x]#.

Note:

When #f(x)=g(x)^(h(x))#, you'll almost always see #g(x)^(h(x))# appear in the derivative of #f(x)#. If you don't, go back and double check your work to make sure things were done right.

Given: #y = x^(sin(x))#

Use logarithmic differentiation.

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

On the right side, use a property of logarithms, #ln(a^b) = (b)ln(a)#:

#ln(y) = (sin(x))ln(x)#

Use implicit differentiation on the left side:

#(dln(y))/dx = 1/ydy/dx#

Use the product rule on the right sides:

#(d(uv))/dx = (u')(v) + (u)(v')#

let #u = sin(x)#, then #u' = cos(x), v =ln(x), and v' = 1/x#

Substituting into the product rule:

#(d((sin(x))ln(x)))/dx = (cos(x))ln(x) + (sin(x))/x#

Put the equation back together:

#1/ydy/dx = (cos(x))ln(x) + (sin(x))/x#

Multiply both sides by y:

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

Substitute #x^(sin(x))# for y:

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