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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives4 Marks
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.$(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})$
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Answer

Let $\text{u}=(3\sec\text{x}-4\text{cosec}\text{x});\text{v}=(-2\sin\text{x}+5\cos\text{x})$Then, $\text{u}'=3\sec\text{x}\tan\text{x}+4\text{cosec}\text{x}\cot\text{x};\text{v}'=-2\cos\text{x}-5\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})]$
$=(3\sec\text{x}-4\text{cosec}\text{x})(-2\cos\text{x}-5\cos\text{x})+(-2\sin\text{x}+5\cos\text{x})(3\sec\text{x}\tan\text{x}+4\text{cosec}\text{x}\cot\text{x})$
$=-6+15\tan\text{x}+8\cot\text{x}+20-6\tan^2\text{x}-8\cot\text{x}-15\tan\text{x}+20\cot^2\text{x}$
$=-6+20-6(\sec^2\text{x}-1)+20(\text{cosec}^2\text{x}-1)$
$=-6\sec^2\text{x}+20\text{cosec}^2\text{x}$
Alternate method
$\frac{\text{d}}{\text{dx}}[​(3\sec\text{x}-4\text{cosec}\text{x})(-2\sin\text{x}+5\cos\text{x})]$
$=\frac{\text{d}}{\text{dx}}(-6\sec\text{x}\sin\text{x}+15\sec\text{x}\cos\text{x}+8\text{cosecx}\sin\text{x}-20\text{cosecx}\cos\text{x})$
$=\frac{\text{d}}{\text{dx}}\Big(-6\frac{\sin\text{x}}{\cos\text{x}}+15\frac{\cos\text{x}}{\cos\text{x}}+8\frac{\sin\text{x}}{\sin\text{x}}-20\frac{\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\text{d}}{\text{dx}}(-6\tan\text{x}+15+8-20\cot\text{x})$
$=\frac{\text{d}}{\text{dx}}(-6\tan\text{x}-20\cot\text{x}+23)$
$-6\sec^2\text{x}+20\text{cosec}^2\text{x}$

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