Question

For the following exercises, use logarithmic differentiation to find dy/dx.

Answer 1

If
`y=x^(\log_2(x))`, then taking the logarithm of both sides gives

`\ln(y) = \ln\left(x^(\log_2(x))\right) = \log_2(x) \ln(x)`

Differentiate both sides with respect to x :

`(d\ln(y))/(dx) = (d\log_2(x))/(dx)\ln(x) + \log_2(x)(d\ln(x))/(dx)`

`\frac1y (dy)/(dx) = (d\log_2(x))/(dx)\ln(x) + (\log_2(x))/(x)`

Now if
`z=\log_2(x)`, then
`2^z=x`. Rewrite

`2^z = e^(\ln(2^z)) = e^(\ln(2)z)`

Then by the chain rule,

`(d2^z)/(dx) = (dx)/(dx)`

`(de^(\ln(2)z))/(dx) = 1`

`e^(\ln(2)z) \ln(2) (dz)/(dx)= 1`

`(dz)/(dx) = (1)/(e^(\ln(2)z)\ln(2))`

`(dz)/(dx) = (1)/(2^z \ln(2))`

`(d\log_2(x))/(dx) = (1)/(\ln(2)x)`

So we have

`\frac1y (dy)/(dx) = (\ln(x))/(\ln(2)x)+ (\log_2(x))/(x)`

`\frac1y (dy)/(dx) = (\log_2(x))/(x)+ (\log_2(x))/(x)`

`\frac1y (dy)/(dx) = (2\log_2(x))/(x)`

`\frac1y (dy)/(dx) = (\log_2(x^2))/(x)`

`(dy)/(dx) = (y\log_2(x^2))/(x)`

Replace y :

`(dy)/(dx) = (x^(\log_2(x))\log_2(x^2))/(x)`

`(dy)/(dx) = x^(\log_2(x)-1)\log_2(x^2)`