1.07d Second derivatives: d^2y/dx^2 notation

\(y = 2x^2 - \frac{4}{\sqrt{x}} + 1\), \(x > 0\)

1. Find \(\frac{dy}{dx}\), giving each term in its simplest form. [3]
2. Find \(\frac{d^2y}{dx^2}\), giving each term in its simplest form. [2]

It is given that \(f(x) = \frac{6}{x^2} + 2x\).

1. Find \(f'(x)\). [3]
2. Find \(f''(x)\). [2]

Given that \(y = 6x^3 + \frac{4}{\sqrt{x}} + 5x\), find

1. \(\frac{\text{d}y}{\text{d}x}\), [4]
2. \(\frac{\text{d}^2y}{\text{d}x^2}\). [2]

Given that $$y = \frac{x^4 - 3}{2x^2},$$

1. find \(\frac{dy}{dx}\), [4]
2. show that \(\frac{d^2y}{dx^2} = \frac{x^4 - 9}{x^4}\). [2]

Given that $$y = \sqrt{x} - \frac{4}{\sqrt{x}},$$

1. find \(\frac{dy}{dx}\). [3]
2. find \(\frac{d^2y}{dx^2}\). [2]
3. show that $$4x\frac{d^2y}{dx^2} + 4x\frac{dy}{dx} - y = 0.$$ [3]

Given that $$\frac{dy}{dx} = \frac{x^3 - 4}{x^2}, \quad x \neq 0,$$

1. find \(\frac{d^2y}{dx^2}\). [3]

Given also that \(y = 0\) when \(x = -1\),

1. find the value of \(y\) when \(x = 2\). [6]

The equation of a curve is \(y = (x^2 + 1)^8\).

1. Find an expression for \(\frac{dy}{dx}\) and hence show that the only stationary point on the curve is the point for which \(x = 0\). [4]
2. Find an expression for \(\frac{d^2y}{dx^2}\) and hence find the value of \(\frac{d^2y}{dx^2}\) at the stationary point. [5]

Given that \(\frac{dy}{dx} = \frac{1}{x}\) find \(\frac{d^2y}{dx^2}\) Circle your answer. [1 mark] \(-\frac{2}{x^2}\) \(-\frac{1}{x^2}\) \(\frac{1}{x^2}\) \(\frac{2}{x^2}\)

It is given that $$y = 3x^4 + \frac{2}{x} - \frac{x}{4} + 1$$ Find an expression for \(\frac{d^2y}{dx^2}\) [3 marks]

The diagram shows the graph of \(y = f(x)\), where \(f(x)\) is a quadratic function of \(x\). A copy of the diagram is given in the Printed Answer Booklet. \includegraphics{figure_2}

1. On the copy of the diagram in the Printed Answer Booklet, draw a possible graph of the gradient function \(y = f'(x)\). [3]
2. State the gradient of the graph of \(y = f''(x)\). [1]