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Exercise 10.3 · Q26

Q.Differentiate the following: y=x+xy = \sqrt{x+\sqrt{x}}

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Step 1. Let u=x+x=x+x1/2u=x+\sqrt{x}=x+x^{1/2}, so y=u=u1/2y=\sqrt{u}=u^{1/2}.

Step 2. dydu=12u=12x+x\dfrac{dy}{du}=\dfrac{1}{2\sqrt{u}}=\dfrac{1}{2\sqrt{x+\sqrt{x}}}.

Step 3. Differentiate the inner function: dudx=1+12x\dfrac{du}{dx}=1+\dfrac{1}{2\sqrt{x}}.

Step 4. Chain rule: dydx=12x+x(1+12x)\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x+\sqrt{x}}}\left(1+\dfrac{1}{2\sqrt{x}}\right).

Step 5. Combine the inner bracket over a common denominator: 1+12x=2x+12x1+\dfrac{1}{2\sqrt{x}}=\dfrac{2\sqrt{x}+1}{2\sqrt{x}}. …

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