First slide

Derivatives

Question

$If y^{2} = a^{2} \cos^{2} x + b^{2} \sin^{2} x then y + y_{2} =$

Moderate

A

$a^{2} b^{2}$
B

$\frac{a^{2} b^{2}}{y^{3}}$
C

$y^{3}$
D

$- \frac{a^{2} b^{2}}{y^{3}}$

Solution

$y^{2} = a^{2} \cos^{2} x + b^{2} \sin^{2} x \Rightarrow 2 y y_{1} = - a^{2} \sin 2 x + b^{2} \sin 2 x$
$\Rightarrow y y_{2} = (b^{2} - a^{2}) \cos 2 x - \frac{{(b^{2} - a^{2})}^{2} \cdot \sin^{2} 2 x}{4 y^{2}} = \frac{(b^{2} - a^{2})}{y^{2}} [a^{2} \cos^{4} x - b^{2} \sin^{4} x]$
$\begin{array}{l} ∴ y^{3} y_{2} = a^{2} b^{2} \cos^{4} x - b^{4} \sin^{4} x - a^{4} \cos^{4} x + a^{2} b^{2} \sin^{4} x \\ y^{4} + y^{3} y_{2} = a^{2} b^{2} {[\cos^{2} x + \sin^{2} x]}^{2} = a^{2} b^{2} \Rightarrow y^{3} (y + y_{2}) = a^{2} b^{2} \Rightarrow y + y_{2} = \frac{a^{2} b^{2}}{y^{3}} \end{array}$

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