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.