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

Question

Isabella Ackerman · Accepted Answer

# ((lnx^2)/x)''=(2(-3+lnx^2))/x^3# The derivative of the quotient #(u(x))/(v(x))# is determined using Quotient Rule #color(blue)(((U(x))/(V(x)))'=(u'(x)v(x)-v'(x)u(x))/v^2# #color(red)((ln(u(x)))'=((u(x))')/(u(x))# #((lnx^2)/x)'=color(blue)(((lnx^2)'*x-x'(lnx^2))/x^2)# #((lnx^2)/x)'=(color(red)((2x)/x^2)*x-1(lnx^2))/x^2# #((lnx^2)/x)'=((2x^2)/x^2-1(lnx^2))/x^2# #color(purple)(((lnx^2)/x)'=(2-lnx^2)/x^2)# #((lnx^2)/x) '' =((color(purple)(lnx^2)/x)')'# #((lnx^2)/x)''=((color(purple)((lnx^2)/x)')'# #((lnx^2)/x)''=((2-lnx^2)/x^2))'# # ((lnx^2)/x)''=color(blue)(((2-lnx^2)'x^2-(x^2)'(2-lnx^2))/x^4 )# # ((lnx^2)/x)''=((0-(2x)/x^2)x^2-(2x)(2-lnx^2))/x^4 # # ((lnx^2)/x)''=(((-2x)/cancelx^2)cancelx^2-(2x)(2-lnx^2))/x^4 # # ((lnx^2)/x)''=(-2x-(2x)(2-lnx^2))/x^4 # # ((lnx^2)/x)''=(-2x-4x+2xlnx^2)/x^4 # # ((lnx^2)/x)''=(-6x+2xlnx^2)/x^4 # # ((lnx^2)/x)''=2x(-3+lnx^2)/x^4 # # ((lnx^2)/x)''=(2(-3+lnx^2))/x^3#

Axel Alvarez · Answer

undefined To find the first and second derivatives of $ \frac{\ln(x^2)}{x} $, you can use the quotient rule and the chain rule. Let's denote $ u = \ln(x^2) $ and $ v = x $. Then, applying the quotient rule:

1. Find the first derivative:
$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} $$

2. Find the second derivative:
$$ \frac{d^2}{dx^2}\left(\frac{u}{v}\right) = \frac{v \frac{d^2u}{dx^2} - 2\frac{du}{dx} \frac{dv}{dx} - u \frac{d^2v}{dx^2}}{v^2} $$

Now, let's compute the derivatives:

1. First derivative:
$$ \frac{du}{dx} = \frac{1}{x^2}(2x) = \frac{2}{x} $$
$$ \frac{dv}{dx} = 1 $$

$$ \frac{d}{dx}\left(\frac{\ln(x^2)}{x}\right) = \frac{x(2/x) - \ln(x^2)(1)}{x^2} = \frac{2 - 2\ln(x)}{x^2} $$

2. Second derivative:
$$ \frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2} $$
$$ \frac{d^2v}{dx^2} = 0 $$

$$ \frac{d^2}{dx^2}\left(\frac{\ln(x^2)}{x}\right) = \frac{x(-2/x^2) - 2\left(\frac{2}{x}\right)(1) - \ln(x^2)(0)}{x^2} $$
$$ = \frac{-2 + 4/x^2}{x^2} = \frac{-2x^2 + 4}{x^4} $$

So, the first derivative is $ \frac{2 - 2\ln(x)}{x^2} $ and the second derivative is $ \frac{-2x^2 + 4}{x^4} $.