Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (c)/Solution 1

Recall from part $a$ that $f^{\prime }(x)=\frac{e^x(x-2)}{x^3}$. We take the derivative of $f^{\prime }$ with respect to $x$ using the quotient rule. Let

$\begin{array}{rl}g(x)& =e^x(x-2)g^{\prime }(x)=e^x(x-2)+e^x=e^x(x-1)\\ h(x)& =x^3{h}^{\prime }(x)=3x^2\end{array}$

By the quotient rule:

$\begin{array}{rl}{f}^{″}(x)& =\frac{g^{\prime }(x)h(x)-g(x)h^{\prime }(x)}{(h(x){)}^2}\\ & =\frac{e^x(x-1)x^3-3e^x(x-2)x^2}{x^6}\\ & =\frac{e^x{x}^2(x(x-1)-3(x-2))}{x^6}\\ & =\frac{e^x(x^2-4x+6)}{x^4}\end{array}$