Find derivative of simple polynomial (integer powers)

3. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = x ^ { 2 } + 3 x - 2\) The point \(P ( 3,16 )\) lies on the curve.

1. Find the gradient of the tangent to the curve at \(P\). The point \(Q\) with \(x\) coordinate \(3 + h\) also lies on the curve.
2. Find, in terms of \(h\), the gradient of the line \(P Q\). Write your answer in simplest form.
3. Explain briefly the relationship between the answer to (b) and the answer to (a).

7. \begin{figure}[h] \end{figure} Figure 4 shows part of the curve with equation \(y = 2 x ^ { 2 } + 5\) The point \(P ( 2,13 )\) lies on the curve.

1. Find the gradient of the tangent to the curve at \(P\). The point \(Q\) with \(x\) coordinate \(2 + h\) also lies on the curve.
2. Find, in terms of \(h\), the gradient of the line \(P Q\). Give your answer in simplest form.
3. Explain briefly the relationship between the answer to (b) and the answer to (a).

4. Given that \(y = 2 x ^ { 5 } + 7 + \frac { 1 } { x ^ { 3 } } , x \neq 0\), find, in their simplest form,

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

4. Given that \(y = 2 x ^ { 2 } - \frac { 6 } { x ^ { 3 } } , x \neq 0\),

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

Given that \(y = x ^ { 4 } + x ^ { \frac { 1 } { 3 } } + 3\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).

4. $$\mathrm { f } ( x ) = 3 x + x ^ { 3 } , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). Given that \(\mathrm { f } ^ { \prime } ( x ) = 15\),
2. find the value of \(x\).

3. Given that \(y = 2 x ^ { 3 } + \frac { 3 } { x ^ { 2 } } , x \neq 0\), find

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. \(\int y \mathrm {~d} x\), simplifying each term.

Given \(y = x ^ { 3 } + 4 x + 1\), find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 3\)

Given that \(y = 4 x ^ { 3 } - \frac { 5 } { x ^ { 2 } } , x \neq 0\), find in their simplest form

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

Given that \(y = 2 x ^ { 5 } + 7 + \frac { 1 } { x ^ { 3 } } , x \neq 0\), find, in their simplest form, (a) \(\frac { \mathrm { d } y } { \mathrm {~d} x }\),
(b) \(\int y \mathrm {~d} x\).

1 Differentiate \(10 x ^ { 4 } + 12\).

1 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - 5 x\).
Find the equation of the curve given that it passes through the point \(( 0,1 )\).

6 \includegraphics[max width=\textwidth, alt={}, center]{918d83c3-1608-4482-9d3d-8af05e65f353-2_394_846_1868_648} The diagram shows part of the curve \(y = x ^ { 2 } + 5\). The point \(A\) has coordinates ( 1,6 ). The point \(B\) has coordinates ( \(a , a ^ { 2 } + 5\) ), where \(a\) is a constant greater than 1 . The point \(C\) is on the curve between \(A\) and \(B\).

1. Find by differentiation the value of the gradient of the curve at the point \(A\).
2. The line segment joining the points \(A\) and \(B\) has gradient 2.3. Find the value of \(a\).
3. State a possible value for the gradient of the line segment joining the points \(A\) and \(C\).

5. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = 3 x ^ { 2 } - 2\) The point \(P ( 2,10 )\) lies on the curve.

1. Find the gradient of the tangent to the curve at \(P\). The point \(Q\) with \(x\) coordinate \(2 + h\) also lies on the curve.
2. Find the gradient of the line \(P Q\), giving your answer in terms of \(h\) in simplest form.
3. Explain briefly the relationship between part (b) and the answer to part (a).

2 Given that \(y = 2 x ^ { 3 }\) find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) Circle your answer.
[0pt] [1 mark] \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 5 x ^ { 2 }\) \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { 2 }\) \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 4 } } { 2 }\) \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { 3 }\)

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

Find \(\frac{dy}{dx}\) in each of the following cases:

1. \(y = \frac{(3x)^2 \times x^4}{x}\), [3]
2. \(y = ^3\sqrt{x}\), [3]
3. \(y = \frac{1}{2x^3}\). [2]

The points \(A(1, 3)\) and \(B(4, 21)\) lie on the curve \(y = x^2 + x + 1\).

1. Find the gradient of the line \(AB\). [2]
2. Find the gradient of the curve \(y = x^2 + x + 1\) at the point where \(x = 3\). [2]

Find \(\frac{dy}{dx}\) when

1. \(y = x - 2x^2\), [2]
2. \(y = \frac{3}{x^2}\). [2]

A and B are points on the curve \(y = 4\sqrt{x}\). Point A has coordinates \((9, 12)\) and point B has \(x\)-coordinate \(9.5\). Find the gradient of the chord AB. The gradient of AB is an approximation to the gradient of the curve at A. State the \(x\)-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]

In Fig. 5, A and B are the points on the curve \(y = 2^x\) with \(x\)-coordinates 3 and 3.1 respectively. \includegraphics{figure_5}

1. Find the gradient of the chord AB. Give your answer correct to 2 decimal places. [2]
2. Stating the points you use, find the gradient of another chord which will give a closer approximation to the gradient of the tangent to \(y = 2^x\) at A. [2]

Find \(\frac{dy}{dx}\) when

1. \(y = 2x^{-5}\). [2]
2. \(y = ^4\sqrt{x}\). [3]

\includegraphics{figure_5} Fig. 5 shows the graph of \(y = 2^x\).

1. On the copy of Fig. 5, draw by eye a tangent to the curve at the point where \(x = 2\). Hence find an estimate of the gradient of \(y = 2^x\) when \(x = 2\). [3]
2. Calculate the \(y\)-values on the curve when \(x = 1.8\) and \(x = 2.2\). Hence calculate another approximation to the gradient of \(y = 2^x\) when \(x = 2\). [2]

The points P\((2, 3.6)\) and Q\((2.2, 2.4)\) lie on the curve \(y = f(x)\). Use P and Q to estimate the gradient of the curve at the point where \(x = 2\). [2]

A and B are points on the curve \(y = 4\sqrt{x}\). Point A has coordinates \((9, 12)\) and point B has \(x\)-coordinate \(9.5\). Find the gradient of the chord AB. The gradient of AB is an approximation to the gradient of the curve at A. State the \(x\)-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]