Tan -1x2-y2-aEvaluate d y-d x

We have, tan^ 1(x^2+y^2)=a On differentiating both sides w.r.t.x. we get d/dxtan^ 1(x^2+y^2)=d/dx(a) ⇒1/1+(x^2+y^2)^2.d/dx(x^2+y^2)=0 ⇒d/dx(x^2+y^2)=0 ⇒ 2x + ...

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Byju's Answer

Standard XII

Mathematics

Average Rate of Change

tan -1x2+y2=a...

Question

$t a n^{- 1} (x^{2} + y^{2}) = a$

Evaluate $\frac{d y}{d x}$

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Solution

We have, $t a n^{- 1} (x^{2} + y^{2}) = a$

On differentiating both sides w.r.t.x. we get

$\frac{d}{d x} t a n^{- 1} (x^{2} + y^{2}) = \frac{d}{d x} (a) \Rightarrow \frac{1}{1 + (x^{2} + y^{2})^{2}} . \frac{d}{d x} (x^{2} + y^{2}) = 0 \Rightarrow \frac{d}{d x} (x^{2} + y^{2}) = 0 \Rightarrow 2 x + 2 y . \frac{d y}{d x} = 0 ∴ \frac{d y}{d x} = \frac{- 2 x}{2 y} = \frac{- x}{y}$

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tan−1(x2+y2)=a Evaluate dydx

We have, tan−1(x2+y2)=a On differentiating both sides w.r.t.x. we get ddxtan−1(x2+y2)=ddx(a)⇒11+(x2+y2)2.ddx(x2+y2)=0⇒ddx(x2+y2)=0⇒2x+2y.dydx=0∴ dydx=−2x2y=−xy

$t a n^{- 1} (x^{2} + y^{2}) = a$

Evaluate $\frac{d y}{d x}$