Question

Find $\frac{dy}{dx}$ if $y={[x+\sqrt{x+\sqrt{x}}]}^{1/2}.$

• A. $\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}}))$
• B. $\frac{dy}{dx}=\frac{1}{2[x+\sqrt{(x+\sqrt{x})}]}$
• C. $\frac{dy}{dx}=\frac{-1}{2\sqrt{[x+\sqrt{(x+\sqrt{x})}]}}$
• D. $\frac{dy}{dx}=\frac{1}{2\sqrt{[x+(x+\sqrt{x})]}}$

Answer 1

The correct option is A $\frac{dy}{dx}=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}}))$

Given

$y={[x+\sqrt{x+\sqrt{x}}]}^{\frac{1}{2}}$

Differentiating $y$ w.r.t $x$, we get

$\frac{dy}{dx}=\frac{1}{2}{[x+\sqrt{x+\sqrt{x}}]}^{-\frac{1}{2}}\frac{d}{dx}(x+\sqrt{x+\sqrt{x}})$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}{[x+\sqrt{x+\sqrt{x}}]}^{-\frac{1}{2}}(1+\frac{1}{2}(x+\sqrt{x}{)}^{-\frac{1}{2}})\frac{d}{dx}(x+\sqrt{x})$

$\Rightarrow \frac{dy}{dx}=\frac{1}{2}{[x+\sqrt{x+\sqrt{x}}]}^{-\frac{1}{2}}(1+\frac{1}{2}(x+\sqrt{x}{)}^{-\frac{1}{2}})(1+\frac{1}{2\sqrt{x}})$