Find second derivative

2. $$y = 2 x ^ { 2 } - \frac { 4 } { \sqrt { } x } + 1 , \quad x > 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), giving each term in its simplest form.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), giving each term in its simplest form. \includegraphics[max width=\textwidth, alt={}, center]{6081d81b-51d2-4140-9834-71ef7fd700b0-05_104_97_2613_1784}

4. $$y = 5 x ^ { 3 } - 6 x ^ { \frac { 4 } { 3 } } + 2 x - 3$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) giving each term in its simplest form.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)

1. $$f ( x ) = x ^ { 3 } + 3 x ^ { 2 } + 5$$ Find

1. \(\mathrm { f } ^ { \prime \prime } ( x )\),
2. \(\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x\).

Given that \(y = 3 x ^ { 2 } + 4 \sqrt { } x , x > 0\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\),
3. \(\int y \mathrm {~d} x\).

3 Given that \(y = 3 x ^ { 5 } - \sqrt { x } + 15\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

6 Given that \(\mathrm { f } ( x ) = ( x + 1 ) ^ { 2 } ( 3 x - 4 )\),

1. express \(\mathrm { f } ( x )\) in the form \(a x ^ { 3 } + b x ^ { 2 } + c x + d\),
2. find \(\mathrm { f } ^ { \prime } ( x )\),
3. find \(\mathrm { f } ^ { \prime \prime } ( x )\).

1. Given that

$$y = \sqrt { x } - \frac { 4 } { \sqrt { x } }$$

1. find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\),
3. show that $$4 x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } - y = 0 .$$

2 Given that \(y = 6 x ^ { \frac { 3 } { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
Show, without using a calculator, that when \(x = 36\) the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) is \(\frac { 3 } { 4 }\).

10 Given tha \(y = 6 x ^ { \frac { 3 } { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
Show, without using a calculator, that when \(x = 36\) the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) is \(\frac { 3 } { 4 }\).

12 Given tha \(y = 6 x ^ { 3 } + \sqrt { x } + 3\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

6 Given that \(y = \frac { 5 } { x ^ { 2 } } - \frac { 1 } { 4 x } + x\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
2. \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).

1 Given that \(\mathrm { f } ( x ) = 6 x ^ { 3 } - 5 x\), find

1. \(\mathrm { f } ^ { \prime } ( x )\),
2. \(f ^ { \prime \prime } ( 2 )\).

The equation of a curve is \(y = kx^{\frac{1}{2}} - 4x^2 + 2\), where \(k\) is a constant.

1. Find \(\frac{\text{d}y}{\text{d}x}\) and \(\frac{\text{d}^2y}{\text{d}x^2}\) in terms of \(k\). [2]
2. It is given that \(k = 2\). Find the coordinates of the stationary point and determine its nature. [4]
3. Points \(A\) and \(B\) on the curve have \(x\)-coordinates 0.25 and 1 respectively. For a different value of \(k\), the tangents to the curve at the points \(A\) and \(B\) meet at a point with \(x\)-coordinate 0.6. Find this value of \(k\). [6]

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

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]

\includegraphics{figure_11} Fig. 11 shows a sketch of the curve with equation \(y = x - \frac{4}{x^2}\).

1. Find \(\frac{dy}{dx}\) and show that \(\frac{d^2y}{dx^2} = -\frac{24}{x^4}\). [3]
2. Hence find the coordinates of the stationary point on the curve. Verify that the stationary point is a maximum. [5]
3. Find the equation of the normal to the curve when \(x = -1\). Give your answer in the form \(ax + by + c = 0\). [5]

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]

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]

A curve has equation \(y = xe^{\frac{x}{2}}\) Show that the curve has a single point of inflection and state the exact coordinates of this point of inflection. [8 marks]