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If   x = \csc \theta - \sin \theta ; y = \csc ^n \theta - \sin ^ n \theta , then \: \: ( x^2 + 4 ) \left ( \frac{dy}{dx} \right )^2 -n ^ 2 y^2 equals to

Option: 1

n ^ 2


Option: 2

2 n^2


Option: 3

3 n^2


Option: 4

4 n^ 2


Answers (1)

best_answer

 

Quotient Rule of differentiation -

\frac{d}{dx}\:\:\frac{f(x)}{g(x)}=\frac{g(x).f'(x)-f(x).g'(x)}{[g(x)]^{2}}


\frac{d}{dx}\:\:\frac{u}{v}=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^{2}}

-

 

 

Derivative of parametric function -

Let\:\:x=f(t)orf(\theta )

y=f(t)\:or\:f(\theta )

Then

\frac{dy}{dx}=\frac{dy}{dt}\div \frac{dx}{dt}\:\:\:or\:\:\:\frac{dy}{d\theta }\div \frac{dx}{d\theta }

- wherein

Let\:\:y=sint

x=cost

\frac{dy}{dt}=cos\:t

\frac{dx}{dt}=-sin\:t

\Rightarrow \:\frac{dy}{dx}=-\frac{cost}{sint}

=\frac{-x}{y}

 

 

\therefore x = \csc \theta - \sin \theta \\\\ \Rightarrow x^2 + 4 = ( \csc \theta + \sin \theta )^2 \\\\ y^2 + 4 = ( \csc^n \theta + \sin ^n\theta )^2 \\\\ now \: \: dy/dx = \frac{dy/d \theta }{dx /d\theta } = \frac{n \cot \theta ( \csc ^ n \theta + \sin ^ n \theta )}{\cot \theta ( \csc \theta + \sin \theta ) } \\\\ \frac{n \sqrt{y^2 + 4 }}{\sqrt x^2 + 4 }

Squaring both sides, we get                                               \left ( \frac{dy}{dx } \right ) ^2 = \frac{n^{2}(y^{2}+4)}{x^{2}+4}\: or \: \left ( x^2 +4 \right ) \left ( \frac{dy}{dx } \right ) ^2 = n^2 ( y ^2 + 4 )

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Riya

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