MHT CET · Maths · Basic of Mathematics
If \(y=\frac{x^{\frac{2}{3}}-x^{\frac{-1}{3}}}{x^{\frac{2}{3}}+x^{\frac{-1}{3}}}, x \neq 0\), then \((x+1)^2 y_1=\)
- A 2
- B -2
- C \(\frac{-1}{3}\)
- D 3
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(y =\frac{x^{\frac{2}{3}}-x^{-\frac{1}{3}}}{x^{\frac{2}{3}}+x^{-\frac{1}{3}}} \)
\( =\frac{x^{-\frac{1}{3}}(x-1)}{x^{-\frac{1}{3}}(x+1)} \)
\( \therefore y =\frac{x-1}{x+1} \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{(x+1) \cdot 1-(x-1) \cdot 1}{(x+1)^2} \)
\( \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{2}{(x+1)^2} \)
\( \Rightarrow(x+1)^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\)
\( =\frac{x^{-\frac{1}{3}}(x-1)}{x^{-\frac{1}{3}}(x+1)} \)
\( \therefore y =\frac{x-1}{x+1} \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{(x+1) \cdot 1-(x-1) \cdot 1}{(x+1)^2} \)
\( \Rightarrow \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{2}{(x+1)^2} \)
\( \Rightarrow(x+1)^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}=2\)
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