How do you find the first and second derivative of (lnx)/x^2lnxx2?

1 Answer
Mar 3, 2018

d/dxlnx/x^2=(1-2lnx)/x^3ddxlnxx2=12lnxx3 and d^2/dx^2lnx/x^2=(6lnx-5)/x^4d2dx2lnxx2=6lnx5x4

Explanation:

Let's rewrite lnx/x^2lnxx2 as x^-2lnx.x2lnx. If we have a positive exponent in the denominator, we can move the term up to the numerator and in turn make the exponent negative.

It's nice to avoid using the Quotient Rule where unnecessary. It saves time and work.

Using the Product Rule:

d/dxlnx/x^2=lnx*d/dx(x^-2)+d/dx(lnx)*x^-2ddxlnxx2=lnxddx(x2)+ddx(lnx)x2

d/dxx^-2=-2x^-3ddxx2=2x3 (Power Rule)

d/dxlnx=1/xddxlnx=1x

Thus,

d/dxx^-2lnx=x^-2(1/x)+(-2x^-3)lnx=x^-2/x-(2lnx)/x^3ddxx2lnx=x2(1x)+(2x3)lnx=x2x2lnxx3

Simplify further:

x^-2/x-(2lnx)/x^3=x^-3-(2lnx)/x^3=1/x^3-(2lnx)/x^3=(1-2lnx)/x^3x2x2lnxx3=x32lnxx3=1x32lnxx3=12lnxx3

For the second derivative, we'll use the quotient rule:

d/dx(1-2lnx)/x^3=(x^3(d/dx1-2lnx)-(1-2lnx)d/dxx^3)/(x^3)^2ddx12lnxx3=x3(ddx12lnx)(12lnx)ddxx3(x3)2

d/dx(1-2lnx)/x^3=(((-2x^3)/x)-3x^2(1-2lnx))/x^6=(-2x^2-3x^2+6x^2lnx)/x^6=(-5x^2+6x^2lnx)/x^6=(x^2(-5+6lnx))/x^6=(6lnx-5)/x^4ddx12lnxx3=(2x3x)3x2(12lnx)x6=2x23x2+6x2lnxx6=5x2+6x2lnxx6=x2(5+6lnx)x6=6lnx5x4