Find the derivative using logarithmic differentiation?

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Answer 1 · 2017-10-17T23:58:43

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

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)#