If \( y=e^{\pi x} \) and \( t=x^{\pi} \), then which of following statements is/are TRUE?
(A) \( \displaystyle\frac{d^2y}{dt^2}=\displaystyle\frac{y}{t^2}\left(\displaystyle\frac{\pi(x-1)+1}{\pi}\right)x \)
(B) \( \displaystyle\frac{d^2y}{dt^2}=y\left(\displaystyle\frac{\pi(x-1)+1}{\pi}\right)x \)
(C) \( \displaystyle\frac{d^2y}{dt^2}=1 \) at \( x=1 \)
(D) \( \displaystyle\frac{d^2y}{dt^2}=\displaystyle\frac{e^{\pi}}{\pi} \) at \( x=1 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
Both y and t are functions of x, so x acts as the parameter.
\( \displaystyle \frac{dy}{dx}=\pi e^{\pi x} \)
\( \displaystyle \frac{dt}{dx}=\pi x^{\pi-1} \)
\( \displaystyle \frac{dy}{dt}=\frac{\pi e^{\pi x}}{\pi x^{\pi-1}} \)
\( \displaystyle =e^{\pi x}\,x^{1-\pi} \)
For the second derivative, \( \frac{dy}{dt} \) is differentiated with respect to x and divided by \( \frac{dt}{dx} \):
\( \displaystyle \frac{d^2y}{dt^2}=\frac{\frac{d}{dx}\left(\frac{dy}{dt}\right)}{\frac{dt}{dx}} \)
By the product rule,
\( \displaystyle \frac{d}{dx}\left(e^{\pi x}x^{1-\pi}\right)=\pi e^{\pi x}x^{1-\pi}+e^{\pi x}(1-\pi)x^{-\pi} \)
\( \displaystyle =e^{\pi x}x^{-\pi}\left[\pi x+1-\pi\right] \)
\( \displaystyle =y\,x^{-\pi}\left[\pi(x-1)+1\right] \)
So
\( \displaystyle \frac{d^2y}{dt^2}=\frac{y\,x^{-\pi}\left[\pi(x-1)+1\right]}{\pi x^{\pi-1}} \)
\( \displaystyle =\frac{y\left[\pi(x-1)+1\right]}{\pi\,x^{2\pi-1}} \)
Since \( t^2=x^{2\pi} \), we have \( x^{2\pi-1}=\frac{t^2}{x} \), so
\( \displaystyle \frac{d^2y}{dt^2}=\frac{y}{t^2}\left(\frac{\pi(x-1)+1}{\pi}\right)x \)
At x = 1: \( y=e^{\pi} \), \( t=1 \) and \( \pi(x-1)+1=1 \), so
\( \displaystyle \frac{d^2y}{dt^2}=\frac{e^{\pi}}{1}\left(\frac{1}{\pi}\right)(1) \)
\( \displaystyle =\frac{e^{\pi}}{\pi} \)
So \( \frac{d^2y}{dt^2}=\frac{y}{t^2}\left(\frac{\pi(x-1)+1}{\pi}\right)x \), and its value at x = 1 is \( \frac{e^{\pi}}{\pi} \).
Checking the statements
(A) \( \displaystyle\frac{d^2y}{dt^2}=\displaystyle\frac{y}{t^2}\left(\displaystyle\frac{\pi(x-1)+1}{\pi}\right)x \) — Correct. This is exactly the second derivative.
(B) \( \displaystyle\frac{d^2y}{dt^2}=y\left(\displaystyle\frac{\pi(x-1)+1}{\pi}\right)x \) — Incorrect. The factor \( \frac{1}{t^2} \) is missing; the two forms agree only when t = 1.
(C) \( \displaystyle\frac{d^2y}{dt^2}=1 \) at \( x=1 \) — Incorrect. The value at x = 1 is \( \frac{e^{\pi}}{\pi} \), not 1.
(D) \( \displaystyle\frac{d^2y}{dt^2}=\displaystyle\frac{e^{\pi}}{\pi} \) at \( x=1 \) — Correct. At x = 1, y = \( e^{\pi} \) and t = 1, giving \( \frac{e^{\pi}}{\pi} \).
Tip: \( \frac{d^2y}{dt^2} \) is not \( \frac{d^2y/dx^2}{d^2t/dx^2} \): differentiate \( \frac{dy}{dt} \) with respect to x and then divide by \( \frac{dt}{dx} \) once more.