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Exercise 10.3 · Q28

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

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Step 1. Name the layers from the inside out: let p=xp=\sqrt{x}, q=x+p=x+xq=x+p=x+\sqrt{x}, r=q=x+xr=\sqrt{q}=\sqrt{x+\sqrt{x}}, so y=x+ry=\sqrt{x+r}.

Step 2. Differentiate the innermost layer: dpdx=12x\dfrac{dp}{dx}=\dfrac{1}{2\sqrt{x}}.

Step 3. Differentiate q=x+pq=x+p: dqdx=1+dpdx=1+12x\dfrac{dq}{dx}=1+\dfrac{dp}{dx}=1+\dfrac{1}{2\sqrt{x}}.

Step 4. Differentiate r=qr=\sqrt{q} using the chain rule: drdx=12q⋅dqdx=1+12x2x+x\dfrac{dr}{dx}=\dfrac{1}{2\sqrt{q}}\cdot\dfrac{dq}{dx}=\dfrac{1+\frac{1}{2\sqrt{x}}}{2\sqrt{x+\sqrt{x}}}.

Step 5. Differentiate the outer layer y=x+ry=\sqrt{x+r} using the chain rule: dydx=12x+r(1+drdx)\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x+r}}\left(1+\dfrac{dr}{dx}\right). …

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