Question
Differentiate the following functions by the product rule and the other method and verify that the answer from both the methods is the same.
$(\text{3x}^2+2)^2$
$(\text{3x}^2+2)^2$
Then,
$\text{u}'=\text{6x};\text{v}'=\text{6x}$Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(\text{3x}^2+2)(\text{3x}^2+2)]$
$=(\text{3x}^2+2)(\text{6x})+(\text{3x}^2+2)(\text{6x})$
$=18\text{x}^3+\text{12x}+\text{18x}^3+\text{12x}$
$=\text{36x}^3+\text{24x}$
Alternate method
$\frac{\text{d}}{\text{dx}}\Big[(\text{3x}^2+2)^2\Big]=\frac{\text{d}}{\text{dx}}(\text{9x}^4+\text{12x}^2+4)$
$=\text{36x}^3+\text{24x}$
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