Question

If
\[ x^3-2x^2y^2+5x+y-5=0, \] then at \((1,1)\), \(y''(1)=\)

• A. \(-\dfrac{197}{27}\)
• B. \(\dfrac{125}{31}\)
• C. \(12\)
• D. \(-\dfrac{238}{27}\)

Hint:

In implicit differentiation, remember that \(y\) is a function of \(x\). So while differentiating terms like \(x^2y^2\), use both product rule and chain rule.

Answer 1

The Correct Option is D

Step 1: Differentiate the given equation once.
Given,
\[ x^3-2x^2y^2+5x+y-5=0 \]
Differentiating with respect to \(x\),
\[ 3x^2-4xy^2-4x^2yy'+5+y'=0 \]

Step 2: Find \(y'\) at \((1,1)\).
Put \(x=1,\;y=1\):
\[ 3-4-4y'+5+y'=0 \]
\[ 4-3y'=0 \]
\[ y'=\frac43 \]

Step 3: Differentiate again.
Differentiate
\[ 3x^2-4xy^2-4x^2yy'+5+y'=0 \]
We get
\[ 6x-4y^2-16xyy'-4x^2(y')^2-4x^2yy''+y''=0 \]

Step 4: Substitute \(x=1,\;y=1,\;y'=\frac43\).
\[ 6-4-16\left(\frac43\right)-4\left(\frac43\right)^2-4y''+y''=0 \]
\[ 2-\frac{64}{3}-\frac{64}{9}-3y''=0 \]
\[ \frac{18-192-64}{9}-3y''=0 \]
\[ -\frac{238}{9}-3y''=0 \]
\[ y''=-\frac{238}{27} \]

Step 5: Final conclusion.
Hence,
\[ \boxed{-\frac{238}{27}} \]