Question

Differentiate X^2/a^2+y^2/b^2=100,where a and b are constants

Answer 1

$\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 100$

We have to differentiate the above expression.

$\dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - 100 = 0$

$\implies\dfrac{d( \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } - 100)}{dx} = \dfrac{d(0)}{dx}$

$\implies \: \dfrac{d(\dfrac{ {x}^{2} }{ {a}^{2} })}{dx} + \dfrac{d(\dfrac{ {y}^{2} }{ {b}^{2} } )}{dx} - \dfrac{d(100)}{dx} = 0$

$\implies \: \dfrac{2x}{ {a}^{2} } + \dfrac{2y}{ {b}^{2} } \: \dfrac{dy}{dx} + 0 = 0$

$\implies \: \dfrac{2x}{ {a}^{2} } + \dfrac{2y}{ {b}^{2} } \: \dfrac{dy}{dx} = 0$

Taking out 2 common :-

$\implies \: 2( \dfrac{x}{ {a}^{2} } + \dfrac{y}{ {b}^{2} } \: \dfrac{dy}{dx} )= 0$

$\implies \: ( \dfrac{x}{ {a}^{2} } + \dfrac{y}{ {b}^{2} } \: \dfrac{dy}{dx} )= \dfrac{0}{2}$

$\implies \: ( \dfrac{x}{ {a}^{2} } + \dfrac{y}{ {b}^{2} } \: \dfrac{dy}{dx} )= {0}$

$\implies \: \dfrac{y}{ {b}^{2} } \: \dfrac{dy}{dx} = - \dfrac{x}{ {a}^{2} }$

$\implies \: \dfrac{dy}{dx} = - \dfrac{x}{ {a}^{2} } \times \dfrac{{b}^{2}}{ y }$

$\implies \: \dfrac{dy}{dx} = - \: \dfrac{x \: {b}^{2}}{ {a}^{2} y}$

Answer :

$\dfrac{dy}{dx} = - \: \dfrac{x \: {b}^{2}}{ {a}^{2} y}$

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$\large\bf\blue{Additional-Information}$

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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