If tan -1x2-y2- then d y-d x is equal to

The correct option is A x/y nbsp; nbsp;tan^ 1(x^2+y^2)=α, α is constant nbsp; ⇒ x^2+y^2= tanα Differentiating wrt x, we get nbsp; 2x+2y dy/dx=0 ⇒dy/ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If tan -1x2+y...

Question

If ${tan}^{- 1} (x^{2} + y^{2}) = α,$ then $\frac{d y}{d x}$ is equal to

\frac{- x}{y}

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x y

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\frac{x}{y}

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- x y

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Solution

The correct option is A $\frac{- x}{y}$
${tan}^{- 1} (x^{2} + y^{2}) = α, α$ is constant
$\Rightarrow x^{2} + y^{2} = tan α$
Differentiating wrt $x$ , we get
$2 x + 2 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = \frac{- x}{y}$

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If tan−1(x2+y2)=α, then dydx is equal to

The correct option is A −xy tan−1(x2+y2)=α, α is constant ⇒x2+y2=tanα Differentiating wrt x, we get 2x+2ydydx=0 ⇒dydx=−xy

If ${tan}^{- 1} (x^{2} + y^{2}) = α,$ then $\frac{d y}{d x}$ is equal to

The correct option is A $\frac{- x}{y}$
${tan}^{- 1} (x^{2} + y^{2}) = α, α$ is constant
$\Rightarrow x^{2} + y^{2} = tan α$
Differentiating wrt $x$ , we get
$2 x + 2 y \frac{d y}{d x} = 0$
$\Rightarrow \frac{d y}{d x} = \frac{- x}{y}$