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Exercise 10.4 · Q7

Q.Find the derivative of the following: x2+y2=tan⁡−1 ⁣(yx)\sqrt{x^2+y^2} = \tan^{-1}\!\left(\dfrac{y}{x}\right)

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Step 1. Let R=x2+y2R=\sqrt{x^2+y^2}. Differentiate the left side:

ddx(R)=x+ydydxR\dfrac{d}{dx}(R) = \dfrac{x+y\frac{dy}{dx}}{R}

Step 2. Differentiate the right side, tan⁡−1(y/x)\tan^{-1}(y/x), using the standard formula:

ddx[tan⁡−1yx]=11+(y/x)2⋅xdydx−yx2=xdydx−yx2+y2=xdydx−yR2\dfrac{d}{dx}\left[\tan^{-1}\dfrac{y}{x}\right] = \dfrac{1}{1+(y/x)^2}\cdot\dfrac{x\frac{dy}{dx}-y}{x^2} = \dfrac{x\frac{dy}{dx}-y}{x^2+y^2} = \dfrac{x\frac{dy}{dx}-y}{R^2}

Step 3. Equate the two derivatives (since R=tan⁡−1(y/x)R = \tan^{-1}(y/x) for all xx):

x+ydydxR=xdydx−yR2\dfrac{x+y\frac{dy}{dx}}{R} = \dfrac{x\frac{dy}{dx}-y}{R^2}

Step 4. Multiply both sides by R2R^2:

R(x+ydydx)=xdydx−yR\left(x+y\dfrac{dy}{dx}\right) = x\dfrac{dy}{dx}-y

Step 5. Expand and collect dydx\dfrac{dy}{dx} terms: …

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