If $y=\frac{x^{\frac{2}{3}}-x^{\frac{-1}{3}}}{x^{\frac{2}{3}}+x^{\frac{-1}{3}}}, x
eq 0$, then $(x+1)^2 y_1=$

Question

If $y=\frac{x^{\frac{2}{3}}-x^{\frac{-1}{3}}}{x^{\frac{2}{3}}+x^{\frac{-1}{3}}}, x 
eq 0$, then $(x+1)^2 y_1=$

Aicharya · Accepted Answer

$\begin{aligned} y & =\frac{x^{\frac{2}{3}}-x^{-\frac{1}{3}}}{x^{\frac{2}{3}}+x^{-\frac{1}{3}}} \ & =\frac{x^{-\frac{1}{3}}(x-1)}{x^{-\frac{1}{3}}(x+1)} \ 	herefore \quad y & =\frac{x-1}{x+1} \ 	herefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{(x+1) \cdot 1-(x-1) \cdot 1}{(x+1)^2} \ & \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{2}{(x+1)^2} \ & \Rightarrow(x+1)^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\end{aligned}$

If $y=\frac{x^{\frac{2}{3}}-x^{\frac{-1}{3}}}{x^{\frac{2}{3}}+x^{\frac{-1}{3}}}, x \neq 0$, then $(x+1)^2…

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