To determine

To find:

The derivative of (sinx)lnx with the use of logarithmic function.

Answer

y′=(sinx)lnx(ln(sinx)x+lnxcotx).

Explanation

Given:

y=(sinx)lnx

Formulae used:

loga(xy)=logax+logayloga(xy)=logax−logaylogaxn=nlogax

Consider y=(sinx)lnx

Differentiate both the sides implicitly with respect to x where ddx(lny)=1yy′.

Finally isolate y′ by multiplying y to both sides and replace yin terms of x and simplify.

lny=ln(sinx)lnxlny=lnx×ln(sinx)1yy′=1x×ln(sinx)+lnx×1sinx×cosxy′=y(ln(sinx)x+lnx×cosxsinx)y′=(sinx)lnx(ln(sinx)x+lnx×cosxsinx)y′=(sinx)lnx(ln(sinx)x+lnxcotx)

Conclusion:

Hence, the derivative of (sinx)1x is 1xcotx−1x2logsinx.