Science:Math Exam Resources/Courses/MATH110/December 2013/Question 07/Solution 1

We can solve this problem by using logarithmic differentiation.

$\begin{array}{rl}\ln (f(x))& =\ln ((x^2+1{)}^7(x^4+2{)}^5(x^6+3{)}^3(x^8+4))\\ & =7\ln (x^2+1)+5\ln (x^4+2)+3\ln (x^6+3)+\ln (x^8+4)\end{array}$

Differentiating implicitly yields

$\begin{array}{rl}\frac{1}{f(x)}{f}^{\prime }(x)& =\frac{7}{x^2+1}\cdot 2x+\frac{5}{x^4+2}\cdot 4x^3+\frac{3}{x^6+3}\cdot 6x^5+\frac{1}{x^8+4}\cdot 8x^7\end{array}$

So

$\begin{array}{rl}{f}^{\prime }(x)& =(x^2+1{)}^7(x^4+2{)}^5(x^6+3{)}^3(x^8+4)(\frac{14x}{x^2+1}+\frac{20x^3}{x^4+2}+\frac{18x^5}{x^6+3}+\frac{8x^7}{x^8+4})\end{array}$