CBSE BoardEnglish MediumSTD 11 ScienceMathsDerivatives2 Marks
Question
Differentiate the following function with respect to $\text{x}:$$(\text{x}^3+\text{x}^2+1)\sin\text{x}$
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Answer
Let $\text{u}=\text{x}^3+\text{x}^2+1;\text{v}=\sin\text{x}$Then, $\text{u}'=\text{3x}^2+\text{2x};\text{v}'=\cos\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(\text{x}^3+\text{x}^2+1)\sin\text{x}]$
$=(\text{x}^3+\text{x}^2+1)\cos\text{x}+(\text{3x}^2+\text{2x})\sin\text{x}$
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