TS EAMCET · Maths · Differentiation
If \(2 x^2-3 x y+y^2+x+2 y-8=0\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{3 y-4 x-1}{2 y-3 x+2}\)
- B \(\frac{3 y+4 x+1}{2 y+3 x+2}\)
- C \(\frac{3 y-4 x+1}{2 y-3 x-2}\)
- D \(\frac{3 y-4 x+1}{2 y+3 x+2}\)
Answer & Solution
Correct Answer
(A) \(\frac{3 y-4 x-1}{2 y-3 x+2}\)
Step-by-step Solution
Detailed explanation
We have, \(2 x^2-3 x y+y^2+x+2 y-8=0\) On differentiating w.r.t. \(x\), we get…
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