Show that derivative equals expression

CAIE P2 2005 November Q7

10 marks Moderate -0.8

7
\includegraphics[max width=\textwidth, alt={}, center]{d527d21f-0ab5-40fa-8cfd-ebfb4aba0a87-3_493_863_264_641} The diagram shows the part of the curve $y = \sin ^ { 2 } x$ for $0 \leqslant x \leqslant \pi$.

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x$.
Hence find the $x$-coordinates of the points on the curve at which the gradient of the curve is 0.5 . [3]
By expressing $\sin ^ { 2 } x$ in terms of $\cos 2 x$, find the area of the region bounded by the curve and the $x$-axis between 0 and $\pi$.

CAIE P3 2021 November Q11

11 marks Standard +0.3

11 The equation of a curve is $y = \sqrt { \tan x }$, for $0 \leqslant x < \frac { 1 } { 2 } \pi$.

Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\tan x$, and verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$ when $x = \frac { 1 } { 4 } \pi$.
The value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ is also 1 at another point on the curve where $x = a$, as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{87be326f-f638-43e9-a654-b7b53d5141ef-18_605_492_1493_822}
Show that $t ^ { 3 } + t ^ { 2 } + 3 t - 1 = 0$, where $t = \tan a$.
Use the iterative formula $$a _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 1 } { 3 } \left( 1 - \tan ^ { 2 } a _ { n } - \tan ^ { 3 } a _ { n } \right) \right)$$ to determine $a$ correct to 2 decimal places, giving the result of each iteration to 4 decimal places.
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Edexcel P3 2023 January Q9

11 marks Standard +0.3

9. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{5abaa077-1da4-4023-b442-194f6972095b-26_659_783_287_641} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows a sketch of part of the curve $C$ with equation $$y = \sqrt { 3 + 4 \mathrm { e } ^ { x ^ { 2 } } } \quad x \geqslant 0$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer in simplest form. The point $P$ with $x$ coordinate $\alpha$ lies on $C$.
Given that the tangent to $C$ at $P$ passes through the origin, as shown in Figure 3,
show that $x = \alpha$ is a solution of the equation $$4 x ^ { 2 } e ^ { x ^ { 2 } } - 4 e ^ { x ^ { 2 } } - 3 = 0$$
Hence show that $\alpha$ lies between 1 and 2
Show that the equation in part (b) can be written in the form $$x = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x ^ { 2 } } }$$ The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \sqrt { 4 + 3 \mathrm { e } ^ { - x _ { n } ^ { 2 } } }$$ with $x _ { 1 } = 1$ is used to find an approximation for $\alpha$.
Use the iteration formula to find, to 4 decimal places, the value of
1. $X _ { 3 }$
2. $\alpha$

OCR MEI C3 2006 January Q1

4 marks Moderate -0.8

1 Given that $y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }$.

OCR MEI C3 2007 June Q1

7 marks Moderate -0.3

1

Differentiate $\sqrt { 1 + 2 x }$.
Show that the derivative of $\ln \left( 1 - \mathrm { e } ^ { - x } \right)$ is $\frac { 1 } { \mathrm { e } ^ { x } - 1 }$.

OCR MEI C3 Q3

4 marks Moderate -0.3

3 Given that $y = \ln \left( \sqrt { \frac { 2 x - 1 } { 2 x + 1 } } \right)$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 x - 1 } - \frac { 1 } { 2 x + 1 }$.

OCR MEI C3 Q5

4 marks Moderate -0.8

5 Given that $y = ( 1 + 6 x ) ^ { \frac { 1 } { 3 } }$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { y ^ { 2 } }$.