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Exercise 10.4 · Q8

Q.Find the derivative of the following: tan⁡(x+y)+tan⁡(x−y)=x\tan(x+y)+\tan(x-y) = x

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Step 1. Given tan⁡(x+y)+tan⁡(x−y)=x\tan(x+y)+\tan(x-y)=x. Differentiate both sides, using the chain rule on each tangent term:

sec⁡2(x+y)(1+dydx)+sec⁡2(x−y)(1−dydx)=1\sec^2(x+y)\left(1+\dfrac{dy}{dx}\right) + \sec^2(x-y)\left(1-\dfrac{dy}{dx}\right) = 1

Step 2. Expand:

sec⁡2(x+y)+sec⁡2(x+y)dydx+sec⁡2(x−y)−sec⁡2(x−y)dydx=1\sec^2(x+y) + \sec^2(x+y)\dfrac{dy}{dx} + \sec^2(x-y) - \sec^2(x-y)\dfrac{dy}{dx} = 1

Step 3. Collect the dydx\dfrac{dy}{dx} terms on one side: …

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