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Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{x}^2(1-\text{x}^2)}{\cos2\text{x}}$
✓

Answer

Consider $\text{y}=\frac{\text{x}^2+2}{\sqrt{\cos\text{x}}}$
Differentiate it with respect to x  and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\cos2\text{x}\frac{\text{d}}{\text{dx}}\text{x}^2(1-\text{x}^2)^3-\text{x}^2(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\cos2\text{x}}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[\text{x}^2\frac{\text{d}}{\text{dx}}(1-\text{x}^2)^3+(1-\text{x}^2)^3\frac{\text{d}}{\text{dx}}\text{x}^2-\text{x}^2(1-\text{x}^2)^3(-2\sin2\text{x})\Big]}{\cos^2 2\text{x}}$
$=\frac{\cos2\text{x}\Big[-6\text{x}^2(1-\text{x}^2)^2+(1-\text{x}^2)^32\text{x}+2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}\Big]}{\cos^2 2\text{x}}$
$=\frac{2\text{x}(1-\text{x}^2)^2}{\cos2\text{x}}-\frac{6\text{x}^3(1-\text{x}^2)^2}{\cos2\text{x}}+\frac{2\text{x}^2(1-\text{x}^2)^3\sin2\text{x}}{\cos^2 2\text{x}}$
$=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=2\text{x}(1-\text{x}^2)\sec2\text{x}\big\{1-4\text{x}^2+\text{x}(1-\text{x}^2)\tan2\text{x}\big\}$

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