If \( y=\left(\sqrt{x}\right)^{\pi} \) such that \( \displaystyle\frac{dy}{dx}=\frac{\pi}{a}y^{\pi+b} \) where \( a,b\in\mathrm{R} \) then:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( a+\pi b=\pi(1-\pi) \)
Since \( y=\left(\sqrt{x}\right)^{\pi}=x^{\pi/2} \), the power rule gives
\( \displaystyle \frac{dy}{dx}=\frac{\pi}{2}x^{\frac{\pi}{2}-1} \)
\( \displaystyle =\frac{\pi}{2}\cdot\frac{x^{\pi/2}}{x} \)
\( \displaystyle =\frac{\pi}{2}\cdot\frac{y}{x} \)
The answer must be in terms of y, so x is expressed through y: \( x=y^{2/\pi} \).
\( \displaystyle \frac{dy}{dx}=\frac{\pi}{2}\cdot\frac{y}{y^{2/\pi}} \)
\( \displaystyle =\frac{\pi}{2}\,y^{1-\frac{2}{\pi}} \)
Comparing with \( \frac{\pi}{a}y^{\pi+b} \):
\( \displaystyle a=2 \)
\( \displaystyle \pi+b=1-\frac{2}{\pi}\ \Rightarrow\ b=1-\frac{2}{\pi}-\pi \)
Then
\( \displaystyle a+\pi b=2+\pi\left(1-\frac{2}{\pi}-\pi\right) \)
\( \displaystyle =2+\pi-2-\pi^{2} \)
\( \displaystyle =\pi(1-\pi) \)