How do you differentiate # y =sqrtln(x^2-3x)# using the chain rule?

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Answer 1 · 2016-01-17T07:01:11

#dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x))#

#y=sqrt(ln (x^2-3x))#

#dy/dx=d/dx(sqrt(ln (x^2-3x)))#

using formula: #d/dx(sqrt(u))=1/(2sqrt(u))*(du)/dx#

#dy/dx=1/(2 sqrt(ln(x^2-3x)))*d/dx(ln(x^2-3x))#

using now the formula:

#d/dx(ln u)=1/u*(du)/dx#

#dy/dx=1/(2 sqrt(ln(x^2-3x)))*1/(x^2-3x)*d/dx(x^2-3x)#

#dy/dx=1/(2 sqrt(ln(x^2-3x)))*1/(x^2-3x)(2x -3)#

After simplification, the final answer is

#dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x))#