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

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MATH110 December 2013

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Question 07
Find the derivative of the function $f(x)=(x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4).$ (There is no need to simplify your answer.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
A question like this is solved best through logarithmic differentiation. However one could do this by using the product rule (see the alternate solution).

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

We can solve this problem by using logarithmic differentiation. ${\begin{aligned}\ln(f(x))&=\ln \left((x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4)\right)\\&=7\ln(x^{2}+1)+5\ln(x^{4}+2)+3\ln(x^{6}+3)+\ln(x^{8}+4)\end{aligned}}$ Differentiating implicitly yields ${\begin{aligned}{\frac {1}{f(x)}}f'(x)&={\frac {7}{x^{2}+1}}\cdot 2x+{\frac {5}{x^{4}+2}}\cdot 4x^{3}+{\frac {3}{x^{6}+3}}\cdot 6x^{5}+{\frac {1}{x^{8}+4}}\cdot 8x^{7}\end{aligned}}$ So ${\begin{aligned}f'(x)&=(x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4)\left({\frac {14x}{x^{2}+1}}+{\frac {20x^{3}}{x^{4}+2}}+{\frac {18x^{5}}{x^{6}+3}}+{\frac {8x^{7}}{x^{8}+4}}\right)\end{aligned}}$

Solution 2

Alternatively, we could solve this problem 'normally' by using product and chain rule. ${\begin{aligned}f'(x)&=7(x^{2}+1)^{6}(x^{4}+2)^{5}(x^{6}+3)^{3}(x^{8}+4)\cdot 2x+5(x^{2}+1)^{7}(x^{4}+2)^{4}(x^{6}+3)^{3}(x^{8}+4)\cdot 4x^{3}\\&\quad +3(x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{2}(x^{8}+4)\cdot 6x^{5}+(x^{2}+1)^{7}(x^{4}+2)^{5}(x^{6}+3)^{3}\cdot 8x^{7}\end{aligned}}$ Note that we made use of the general product rule: ${\begin{aligned}{\frac {d}{dx}}(f_{1}f_{2}f_{3}f_{4})&={\frac {df_{1}}{dx}}f_{2}f_{3}f_{4}+f_{1}{\frac {df_{2}}{dx}}f_{3}f_{4}+f_{1}f_{2}{\frac {df_{3}}{dx}}f_{4}+f_{1}f_{2}f_{3}{\frac {df_{4}}{dx}}.\end{aligned}}$ which can be found by applying the product rule 3 separate times.

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