How do you use the quotient rule to differentiate #(2lnx^3) / sqrt(2-x^2)#?

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Answer 1 · 2018-03-18T11:41:21

#=>(dy)/(dx)=6[(2-x^2+x^2lnx)/(x(2-x^2)^(3/2))]#

Here, #2lnx^3=2(3lnx)=6lnx#

So, let us take

#y=(6lnx)/sqrt(2-x^2)#

Now, #color(red)(d/(dx)(u/v)=(v*u^'-u*v^')/v^2)#

#:.(dy)/(dx)=6[(sqrt(2-x^2)*1/x-lnx*1/(cancel(2)(sqrt(2-x^2)))(-cancel(2)x))/((sqrt(2-x^2))^2)]#

#=6[(sqrt(2-x^2)/x+(xlnx)/(sqrt(2-x^2)))/(2-x^2)]#

#=6[((2-x^2)+(x^2lnx))/(xsqrt(2-x^2)(2-x^2))]#

#=>(dy)/(dx)=6[(2-x^2+x^2lnx)/(x(2-x^2)^(3/2))]#