How do you find the second derivative of #ln(x^2-3x+3)# ?

Question

Elijah Abram · Accepted Answer

Firstly, use implicit differentiation...

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

#e^y=x^2-3x+3#

#e^y*(dy)/(dx)=2x-3#

#(dy)/(dx)=(2x-3)/e^y#

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

Now use the quotient rule...

#(d^2y)/(dx^2)=((x^2-3x+3)*2-(2x-3)(2x-3))/((x^2-3x+3)^2)#

And simplify the fraction...

#(d^2y)/(dx^2)=(2x^2-6x+6-(4x^2-12x+9))/((x^2-3x+3)^2)#

#(d^2y)/(dx^2)=(2x^2-6x+6-4x^2+12x-9)/((x^2-3x+3)^2)#

#(d^2y)/(dx^2)=(-2x^2+6x-3)/((x^2-3x+3)^2)#

#(d^2y)/(dx^2)=(6x-2x^2-3)/((x^2-3x+3)^2)#

Emma Aldridge · Answer

undefined To find the second derivative of $ \ln(x^2 - 3x + 3) $, we first need to find the first derivative and then apply the chain rule twice.

First derivative:
$ \frac{d}{dx}(\ln(x^2 - 3x + 3)) = \frac{1}{x^2 - 3x + 3} \cdot \frac{d}{dx}(x^2 - 3x + 3) $

Second derivative:
$ \frac{d^2}{dx^2}(\ln(x^2 - 3x + 3)) = \frac{d}{dx}\left(\frac{1}{x^2 - 3x + 3}\right) $

To find the derivative of $ \frac{1}{x^2 - 3x + 3} $, we use the quotient rule.

Let $ u = 1 $ and $ v = x^2 - 3x + 3 $.

Then, applying the quotient rule:

$ \frac{d}{dx}\left(\frac{1}{x^2 - 3x + 3}\right) = \frac{v \cdot \frac{d}{dx}(u) - u \cdot \frac{d}{dx}(v)}{v^2} $

$ = \frac{0 \cdot (x^2 - 3x + 3) - 1 \cdot (2x - 3)}{(x^2 - 3x + 3)^2} $

$ = -\frac{2x - 3}{(x^2 - 3x + 3)^2} $

So, the second derivative of $ \ln(x^2 - 3x + 3) $ is $ -\frac{2x - 3}{(x^2 - 3x + 3)^2} $.