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Parametric Differentiation

If t 1+ x 2= ...

Question

If $t (1 + x^{2}) = x$ and $x^{2} + t^{2} = y$ then at x = 2, the value of $\frac{d y}{d x}$ is

A

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B

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C

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D

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Solution

The correct option is C
$\frac{d y}{d x} = 2 x + 2 t . \frac{d t}{d x} t = \frac{x}{1 + x^{2}} \frac{d t}{d x} = \frac{1 - x^{2}}{(1 + x^{2})^{2}} p u t x = 2$

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Similar questions

t (1 + x^{2}) = x

x^{2} + t^{2} = y,

x = 2,

\frac{d y}{d x}

t (1 + x^{2}) = x

x^{2} + t^{2} = y

\frac{d y}{d x}

x = 2

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

\frac{d y}{d x} = a x + b

, then the values of

a

b

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

\frac{d y}{d x}

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

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

, then prove that

\frac{d y}{d x} = - \frac{y}{x}

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