y = x + x + x We apply chain rule while differentiating d y d x = d d x ( x + x + x ) We know that d d x ( g ( x ) ) = 1 2 g ( x ) d d x ( g ( x ) ) So, we have d y d x = d d x ( x + x + x ) = 1 2 x + x + x d d x ( x + x + x ) = 1 2 x + x + x ( 1 + 1 2 x + x d d x ( x + x ) ) = 1 2 x + x + x ( 1 + 1 2 x + x ( 1 + 1 2 x ) ) y=\sqrt{x+\sqrt{x+\sqrt{x}}}\\
\text{We apply chain rule while differentiating}\\
\frac{dy}{dx}=\frac{d}{dx}(\sqrt{x+\sqrt{x+\sqrt{x}}})\\
\text{We know that }\frac{d}{dx}(\sqrt{g(x)})=\frac{1}{2\sqrt{g(x)}} \frac{d}{dx}(g(x))\\
\text{So, we have}\\
\frac{dy}{dx}=\frac{d}{dx}(\sqrt{x+\sqrt{x+\sqrt{x}}})\\
\;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}\frac{d}{dx}(x+\sqrt{x+\sqrt{x}})\\
\;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}\frac{d}{dx}(x+\sqrt{x}))\\
\;\;\;\;\;=\frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}}(1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}}))\\ y = x + x + x We apply chain rule while differentiating d x d y = d x d ( x + x + x ) We know that d x d ( g ( x ) ) = 2 g ( x ) 1 d x d ( g ( x )) So, we have d x d y = d x d ( x + x + x ) = 2 x + x + x 1 d x d ( x + x + x ) = 2 x + x + x 1 ( 1 + 2 x + x 1 d x d ( x + x )) = 2 x + x + x 1 ( 1 + 2 x + x 1 ( 1 + 2 x 1 ))
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