AP EAMCET · Maths · Application of Derivatives
The point which lies on the tangent drawn to the curve \(x^4 e^y+2 \sqrt{y+1}=3\) at the point \((1,0)\) is
- A \((2,6)\)
- B \((2,-6)\)
- C \((-2,-6)\)
- D \((-2,6)\)
Answer & Solution
Correct Answer
(D) \((-2,6)\)
Step-by-step Solution
Detailed explanation
\(x^4 e^y+2 \sqrt{y+1}=3\) Differentiate w.r.t \(x\) \(x^4 e^y y^{\prime}+e^y 4 x^3+\frac{2 y^{\prime}}{2 \sqrt{y+1}}=0\) at \((1,0)\) \(y^{\prime}+4+y^{\prime}=0 \Rightarrow y^{\prime}=-2\) Tangent at \((1,0)\) is \(y-0=-2(x-1) \Rightarrow 2 x+y=2\) \((-2,6)\) lies on it
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