MHT CET · Maths · Differentiation
If \(y^2=a x^2+b x+c\), where \(a, b, c\) are constants, then \(y^3 \frac{d^2 y}{d x^2}\) is equal to
- A functions of \(y\)
- B function of both \(\mathrm{x}\) and \(\mathrm{y}\)
- C constant
- D function of \(x\)
Answer & Solution
Correct Answer
(D) function of \(x\)
Step-by-step Solution
Detailed explanation
\(
y^2=a x^2+b x+c
\)
Differentiating w.r.t. \(x\), we get
\(2 y \frac{d y}{d x}=2 a x+b \Rightarrow 2 y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2=2 a \)
\( \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=a \)
\( \therefore y^3 \frac{d^2 y}{d x^2}=\left(a x^2+b x+c\right)\left[\left(\frac{2 a x+b}{2}\right)^2-a\right]\)
R.H.S. of eq. (1) is a function of ' \(x\) ' only.
y^2=a x^2+b x+c
\)
Differentiating w.r.t. \(x\), we get
\(2 y \frac{d y}{d x}=2 a x+b \Rightarrow 2 y \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^2=2 a \)
\( \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=a \)
\( \therefore y^3 \frac{d^2 y}{d x^2}=\left(a x^2+b x+c\right)\left[\left(\frac{2 a x+b}{2}\right)^2-a\right]\)
R.H.S. of eq. (1) is a function of ' \(x\) ' only.
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