find dy/dx where y=2^(x/lnx)

Question

find dy/dx where y=2^(x/lnx)

Answer 1

To solve this derivative, you need the chain rule and the quotient rule.

 

The chain rule basically states that you need to take the derivative of the outside function and multiply it by the derivative of the inside function: f'(g(x))*g'(x). So we need to establish what the outside function is. The outside function for this equation is f(x) = 2^x, and the inside function is g(x) = x/lnx.

 

The derivative of f(x) = 2^x is f'(x) = 2^xln(2)  <--- Derivative rule: y = a^x --> y' = a^xln(a)

The chain rule shows that we need f'(g(x)): f'(g(x)) = 2^x/lnxln2

 

Now we need to find g'(x) to finish the problem. g(x) = x/lnx is a quotient, so we'll use the quotient rule (and I'm going to use h(x) and j(k) to hopefully alleviate confusion): [h'(x)j(x) - h(x)j'(x)]/[j(x)]². h(x) in this equation is x, and j(x) = lnx. It can be helpful to write out a little list before solving the derivative:

h(x) = x                      j(x) = lnx

h'(x) = 1                     j'(x) = 1/x

Now we just plug these values into the quotient rule and reduce:

[1(lnx) - x(1/x)]/[lnx]²

[lnx -1]/[lnx]²

This is g'(x)

 

So we now just need to finish the chain rule by multiplying f'(g(x)) by g'(x):

2^x/lnxln(2)([lnx - 1]/[lnx]²)

 

y' = [2^x/lnxln(2)(lnx-1)]/(lnx)²