TS EAMCET · Maths · Differentiation
If \(2 x^2+3 x y-y^2+4 x-5 y+6=0\), then the value of \(\frac{d y}{d x}\) at \((x, y)=(1,-2)\) is
- A 1
- B -1
- C \(\frac{7}{2}\)
- D 0
Answer & Solution
Correct Answer
(B) -1
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} 2 x^2+3 x y-y^2+4 x-5 y+6=0 \\ & \Rightarrow \quad 2(2 x)+3\left\{x \cdot \frac{d y}{d x}+y\right\}-2 y \frac{d y}{d x}+4-5 \frac{d y}{d x}=0 \\ & \Rightarrow \quad(4 x+3 y+4)+(3 x-2 y-5) \frac{d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d…
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