Differentiate with respect to x from the first principle?

#f(x)#=#x^(2sinx)#

1 Answer
Feb 24, 2018

#f'(x)=x^(2sin(x))((2sin(x))/x+2cos(x)ln(x))#

Explanation:

#y=f(x)#

#y=x^(2sin(x))#

The Power Rule for cannot be used here, because #x# is not raised to a constant power, rather it is raised to the power of another function. Whenever we encounter a function in the exponent, we should apply the natural logarithm to both sides :

#ln(y)=ln(x^(2sin(x)))#

Recall the exponent rule for logarithms, which states that #ln(x^a)=aln(x)#. This rule applies even if #a# is a function and not just a constant.

#ln(y)=2sin(x)ln(x)#

Differentiate both sides with respect to #x#. This means the chain rule will apply for differentiating #ln(y)#, giving us an instance of #dy/dx:#

#1/y * dy/dx=(2sin(x))/x+2ln(x)cos(x)#

Solve for #dy/dx# by multiplying both sides by #y#:

#dy/dx=y((2sin(x))/x+2cos(x)ln(x))#

Recall that #y=x^(2sin(x))#:

#dy/dx=x^(2sin(x))((2sin(x))/x+2cos(x)ln(x))#

#f'(x)=x^(2sin(x))((2sin(x))/x+2cos(x)ln(x))#