Question

If \(\mathrm{a}\left(4+x^2\right)=x\) and \(y-x^3=\mathrm{a}^2\) then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=1\) is \(\qquad\)

• A. \(\frac{441}{125}\)
• B. \(\frac{18}{125}\)
• C. \(\frac{378}{125}\)
• D. \(\frac{381}{125}\)

Hint:

Express a in terms of x, then differentiate y = x^3 + a^2 with respect to x.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
From the first relation \(a=\frac{x}{4+x^2}\), and \(y=x^3+a^2\).

Step 2: Differentiate
\(\frac{dy}{dx}=3x^2+2a\frac{da}{dx}\)
\(\frac{da}{dx}=\frac{(4+x^2)-2x^2}{(4+x^2)^2}=\frac{4-x^2}{(4+x^2)^2}\).

Step 3: At x = 1
\(a=\frac15\) and \(\frac{da}{dx}=\frac{3}{25}\).

Step 4: Substitute
\(\frac{dy}{dx}=3+2\cdot\frac15\cdot\frac{3}{25}=3+\frac{6}{125}=\frac{381}{125}\).

Final Answer:
The derivative is \(\frac{381}{125}\). \[ \boxed{\text{(D) }\dfrac{381}{125}} \]