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).

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Solution 1

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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

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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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