Question

Find the derivative using logarithmic differentiation?

Answer 1

`dy/dx=1/2sqrtx^sinx(sin/x+lnxcosx)`

Explanation:

Take the natural log of both sides.

`lny=lnsqrtx^sinx`

Using one of the properties of logarithms (`loga^b=bloga`), we can rewrite the equations as:

`lny=1/2sinxlnx`

Take the derivative of both sides, using the product rule on the right and implicit differentiation on the left.

`(1/y)dy/dx=1/2(sinx/x+lnxcosx)`

Multiply both sides by `1/y`

`dy/dx=y/2(sinx/x+lnxcosx)`

Finally, substitute the original equation for `y`.

`dy/dx=1/2sqrtx^sinx(sin/x+lnxcosx)`