If ${x^2} + {y^2} = t + \frac{1}{t}$ and ${x^4} + {{\rm{y}}^4} = {{\rm{t}}^2} + \frac{1}{{{{\rm{t}}^2}}},$ then $\frac{{dy}}{{dx}}$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

{x^2} + {y^2} = t + \frac{1}{t}

and

{x^4} + {\mathrm{y}^4} = {\mathrm{t}^2} + \frac{1}{{{\mathrm{t}^2}}},

then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: B

- \frac{y}{x}

Option B (

\ - \frac{y}{x}

) is correct because by manipulating the given equations, we find that

{x^2}{y^2} = 1

, and differentiating this yields

\frac{{dy}}{{dx}} = - \frac{y}{x}

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f(x)=|x|^{|\tan x|}

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y

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[x]

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\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

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v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}