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 relation once.
From $x^3-2x^2y^2+5x+y-5=0$, differentiating term by term gives $3x^2-4xy^2-4x^2y\,y'+5+y'=0$.
Step 2: Get $y'$ at $(1,1)$.
Putting $x=1,y=1$: $3-4-4y'+5+y'=0$, so $4-3y'=0$ and $y'=\dfrac{4}{3}$.
Step 3: Differentiate a second time.
Differentiating $3x^2-4xy^2-4x^2y\,y'+5+y'=0$ gives $6x-4y^2-8xy\,y'-\big(8xy\,y'+4x^2(y')^2+4x^2y\,y''\big)+y''=0$, which tidies to $6x-4y^2-16xy\,y'-4x^2(y')^2-4x^2y\,y''+y''=0$.
Step 4: Substitute the known values.
With $x=1,y=1,y'=\dfrac43$: $6-4-16\cdot\dfrac43-4\cdot\dfrac{16}{9}-4y''+y''=0$.
Step 5: Combine the constants.
$6-4=2$, $16\cdot\dfrac43=\dfrac{64}{3}$, $4\cdot\dfrac{16}{9}=\dfrac{64}{9}$. So $2-\dfrac{64}{3}-\dfrac{64}{9}-3y''=0$, and $2-\dfrac{64}{3}-\dfrac{64}{9}=\dfrac{18-192-64}{9}=-\dfrac{238}{9}$.
Step 6: Solve for $y''$.
$-\dfrac{238}{9}=3y''$, so $y''=-\dfrac{238}{27}$.
\[ \boxed{-\dfrac{238}{27}} \]