If \(y = 2x^{n+1} + \frac{3}{x^n}\), then we need to find \(x^2 \frac{d^2y}{dx^2}\).
Rewrite y as: \(y = 2x^{n+1} + 3x^{-n}\)
First, find the first derivative \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}\)
Next, find the second derivative \(\frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2}\)
\(\frac{d^2y}{dx^2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}\)
Now, multiply by \(x^2\):
\(x^2 \frac{d^2y}{dx^2} = x^2 (2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2})\)
\(x^2 \frac{d^2y}{dx^2} = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n}\)
\(x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + 3x^{-n})\)
\(x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + \frac{3}{x^n})\)
Since \(y = 2x^{n+1} + \frac{3}{x^n}\), we have:
\(x^2 \frac{d^2y}{dx^2} = n(n+1)y\)
Therefore, the correct option is (B) \(n(n+1)y\).
Given $y = 2x^{n+1} + \frac{3}{x^n} = 2x^{n+1} + 3x^{-n}$.
First derivative: $\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}$.
Second derivative: $$\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}$$ $$x^2 \frac{d^2y}{dx^2} = x^2 (2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}) = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n}$$ $$x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + 3x^{-n}) = n(n+1)y$$
A cylindrical tank of radius 10 cm is being filled with sugar at the rate of 100π cm3/s. The rate at which the height of the sugar inside the tank is increasing is:
If \(f(x) = \begin{cases} x^2 + 3x + a, & x \leq 1 bx + 2, & x>1 \end{cases}\), \(x \in \mathbb{R}\), is everywhere differentiable, then