Show derivative equals given algebraic form

9 Fig. 9 shows the curve \(y = \frac { x ^ { 2 } } { 3 x - 1 }\).
P is a turning point, and the curve has a vertical asymptote \(x = a\). \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }\).
3. Find the exact coordinates of the turning point P . Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point.
4. Using the substitution \(u = 3 x - 1\), show that \(\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac { 2 } { 3 }\) and \(x = 1\).

5. Given that $$y = \frac { 5 \cos \theta } { 4 \cos \theta + 4 \sin \theta } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$ Show that $$\frac { d y } { d \theta } = - \frac { 5 } { 4 ( 1 + \sin 2 \theta ) } , \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$

12. A curve has equation \(y = \frac { 2 x e ^ { x } } { x + k }\) where \(k\) is a positive constant.
i. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { e ^ { x } \left( 2 x ^ { 2 } + 2 k x + 2 k \right) } { ( x + k ) ^ { 2 } }\) ii. Given that the curve has exactly one stationary point find the value of \(k\).

1. Given that

$$y = \frac { 3 \sin \theta } { 2 \sin \theta + 2 \cos \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$ show that $$\frac { d y } { d \theta } = \frac { A } { 1 + \sin 2 \theta } \quad - \frac { \pi } { 4 } < \theta < \frac { 3 \pi } { 4 }$$ where \(A\) is a rational constant to be found.

3. $$y = \frac { 5 x ^ { 2 } + 10 x } { ( x + 1 ) ^ { 2 } } \quad x \neq - 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A } { ( x + 1 ) ^ { n } }\) where \(A\) and \(n\) are constants to be found.
2. Hence deduce the range of values for \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } < 0\)

1. A curve has equation \(y = \mathrm { f } ( x )\), where

$$\mathrm { f } ( x ) = \frac { 7 x \mathrm { e } ^ { x } } { \sqrt { \mathrm { e } ^ { 3 x } - 2 } } \quad x > \ln \sqrt [ 3 ] { 2 }$$

1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 7 \mathrm { e } ^ { x } \left( \mathrm { e } ^ { 3 x } ( 2 - x ) + A x + B \right) } { 2 \left( \mathrm { e } ^ { 3 x } - 2 \right) ^ { \frac { 3 } { 2 } } }$$ where \(A\) and \(B\) are constants to be found.
2. Hence show that the \(x\) coordinates of the turning points of the curve are solutions of the equation $$x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$$ The equation \(x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }\) has two positive roots \(\alpha\) and \(\beta\) where \(\beta > \alpha\) A student uses the iteration formula $$x _ { n + 1 } = \frac { 2 \mathrm { e } ^ { 3 x _ { n } } - 4 } { \mathrm { e } ^ { 3 x _ { n } } + 4 }$$ in an attempt to find approximations for \(\alpha\) and \(\beta\) Diagram 1 shows a plot of part of the curve with equation \(y = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }\) and part of the line with equation \(y = x\) Using Diagram 1 on page 42
3. draw a staircase diagram to show that the iteration formula starting with \(x _ { 1 } = 1\) can be used to find an approximation for \(\beta\) Use the iteration formula with \(x _ { 1 } = 1\), to find, to 3 decimal places,
4. 1. the value of \(x _ { 2 }\)
   2. the value of \(\beta\) Using a suitable interval and a suitable function that should be stated
5. show that \(\alpha = 0.432\) to 3 decimal places. Only use the copy of Diagram 1 if you need to redraw your answer to part (c). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_736_812_372_143} \captionsetup{labelformat=empty} \caption{Diagram 1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_738_815_370_1114} \captionsetup{labelformat=empty} \caption{copy of Diagram 1}
   \end{figure}

1. The function f is defined by

$$f ( x ) = \frac { 2 x - 3 } { x ^ { 2 } + 4 } \quad x \in \mathbb { R }$$

1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { a x ^ { 2 } + b x + c } { \left( x ^ { 2 } + 4 \right) ^ { 2 } }$$ where \(a\), \(b\) and \(c\) are constants to be found.
2. Hence, using algebra, find the values of \(x\) for which f is decreasing. You must show each step in your working.

1. Given that

$$y = \frac { x - 4 } { 2 + \sqrt { x } } \quad x > 0$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \mathrm {~A} \sqrt { \mathrm { x } } } \quad x > 0$$ where \(A\) is a constant to be found.

1. The function f is defined by

$$f ( x ) = \frac { e ^ { 3 x } } { 4 x ^ { 2 } + k }$$ where \(k\) is a positive constant.

1. Show that $$f ^ { \prime } ( x ) = \left( 12 x ^ { 2 } - 8 x + 3 k \right) g ( x )$$ where \(\mathrm { g } ( x )\) is a function to be found. Given that the curve with equation \(y = \mathrm { f } ( x )\) has at least one stationary point, (b) find the range of possible values of \(k\).

3. Given \(y = x ( 2 x + 1 ) ^ { 4 }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) ^ { n } ( A x + B )$$ where \(n , A\) and \(B\) are constants to be found.

1

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when:
   1. \(y = \left( 2 x ^ { 2 } - 5 x + 1 \right) ^ { 20 }\);
   2. \(y = x \cos x\).
2. Given that $$y = \frac { x ^ { 3 } } { x - 2 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k x ^ { 2 } ( x - 3 ) } { ( x - 2 ) ^ { 2 } }$$ where \(k\) is a positive integer.

2. (a) Differentiate with respect to \(x\)

1. \(3 \sin ^ { 2 } x + \sec 2 x\),
2. \(\{ x + \ln ( 2 x ) \} ^ { 3 }\). Given that \(y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } , x \neq 1\),
   (b) show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 8 } { ( x - 1 ) ^ { 3 } }\).

1

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = \tan 3 x\).
   (2 marks)
2. Given that \(y = \frac { 3 x + 1 } { 2 x + 1 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { ( 2 x + 1 ) ^ { 2 } }\).
   (3 marks)

9

1. Given that \(y = \frac { 4 x } { 4 x - 3 }\), use the quotient rule to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( 4 x - 3 ) ^ { 2 } }\), where \(k\) is an integer.
2. 1. Given that \(y = x \ln ( 4 x - 3 )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Find an equation of the tangent to the curve \(y = x \ln ( 4 x - 3 )\) at the point where \(x = 1\).
3. 1. Use the substitution \(u = 4 x - 3\) to find \(\int \frac { 4 x } { 4 x - 3 } \mathrm {~d} x\), giving your answer in terms of \(x\).
   2. By using integration by parts, or otherwise, find \(\int \ln ( 4 x - 3 ) \mathrm { d } x\).

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

1. Show that \(\frac{dy}{dx} = \frac{2x(x + 1)}{(2x + 1)^2}\). [4]
2. Find the coordinates of the stationary points of the curve. You need not determine their nature. [4]

The curve \(C\) has equation $$y = \frac{3 + \sin 2x}{2 + \cos 2x}$$

1. Show that $$\frac{dy}{dx} = \frac{6\sin 2x + 4\cos 2x + 2}{(2 + \cos 2x)^2}$$ [4]
2. Find an equation of the tangent to \(C\) at the point on \(C\) where \(x = \frac{\pi}{2}\). Write your answer in the form \(y = ax + b\), where \(a\) and \(b\) are exact constants. [4]