List of top Mathematics Questions on Second Order Derivative

For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:

If \(y(x)=x^x,\ x>0\), then \[ y''(2)-2y'(2)= \]

For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:

If \( y = \cos^{-1} x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \).

The second derivative of \( \sin 3x \cos 5x \) is:

If $y=e^{mx}\sin(nx)$, then the value of \[ \frac{d^2y}{dx^2}-2m\frac{dy}{dx}+(m^2+n^2)y \] is:

If $\log y=\log(\sin x)-x^2$, then \[ \frac{d^2y}{dx^2}+4x\frac{dy}{dx}+4x^2y= \]

If $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$, and it is given that $\frac{d^2y}{dx^2} = 0$, then:

If $f(x)=(2x+3)e^{5x}$, then $f^{\prime\prime}(1)-10f^{\prime}(1)$ is equal to

If $y = e^{-x^2}$, then $\dfrac{d^2y}{dx^2} + 2x\dfrac{dy}{dx}$ is equal to:

If \( y=(5x-2)e^x \), then \( \dfrac{d^2y}{dx^2} \) is equal to

If \( y = x + e^x \) then \( \frac{d^2 x}{dy^2} = \)

If \[ y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2, \] then \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = \]

If \(y = (x - 1)\log_{e}(x - 1)\), then \(\frac{d^{2}y}{dx^{2}}\) at \(x = 3\) is

If $t = e^{2x}$ and $y = \ln(t^2)$, then $\frac{d^2 y}{dx^2}$ is:

The second-order derivative of which of the following functions is $5^x$?

If \[ x = \sqrt{e^{\sin^{-1} t}} \quad \text{and} \quad y = \sqrt{e^{\cos^{-1} t}}, \] then find \[ \frac{d^2 y}{dx^2}. \]

Let $f$ be a twice differentiable function on $R$. 
If $f ^{\prime}(0)=4$ and $f(x)+\int\limits_0^x(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x,$ 
then $(2 a+1)^5 a^2$ is equal to ______.

If $y = \tan^{-1}\left[\frac{\log\left(\frac{e}{x^2}\right)}{\log(ex^2)}\right] + \tan^{-1}\left[\frac{3 + 2\log x}{1 - 6\log x}\right]$, then $\frac{d^2y}{dx^2} = $

If \( y = 2 \sin x + 3 \cos x \) and \( y + A \frac{d^2 y}{dx^2} = B \), then the values of \( A, B \) are respectively

If \( x = A\cos 4t + B\sin 4t \), then \( \frac{d^2x}{dt^2} = \)

If \[ y = \frac{x}{x+1} + \frac{x+1}{x}, \] then \[ \frac{d^2 y}{dx^2} \text{ at } x = 1 \text{ is equal to:} \]

Let \(y = e^{2x}\). Then \(\frac{d^2 y}{dx^2} \cdot \frac{d^2 x}{dy^2}\) is:

If \( y = \frac{x}{x+1} + \frac{x+1}{x} \), then \( \frac{d^2y}{dx^2} \) at \( x=1 \) is equal to