Show derivative satisfies condition

1 The equation of a curve is \(y = \frac { 1 + x } { 1 + 2 x }\) for \(x > - \frac { 1 } { 2 }\). Show that the gradient of the curve is always negative.

2 The equation of a curve is \(y = \frac { \sin x } { 1 + \cos x }\), for \(- \pi < x < \pi\). Show that the gradient of the curve is positive for all \(x\) in the given interval.

11. $$y = \left( 2 x ^ { 2 } - 3 \right) \tan \left( \frac { 1 } { 2 } x \right) , \quad 0 < x < \pi$$

1. Find the exact value of \(x\) when \(y = 0\) Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = \alpha\),
2. show that $$2 \alpha ^ { 2 } - 3 + 4 \alpha \sin \alpha = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 3 } { \left( 2 x _ { n } + 4 \sin x _ { n } \right) }$$ can be used to find an approximation for \(\alpha\).
3. Taking \(x _ { 1 } = 0.7\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 0.7283\) to 4 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{29b56d51-120a-4275-a761-8b8aed7bca54-38_2253_50_314_1977}

2. A curve \(C\) has equation $$y = \mathrm { e } ^ { 2 x } \tan x , \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 }$$

1. Show that the turning points on \(C\) occur where \(\tan x = - 1\).
2. Find an equation of the tangent to \(C\) at the point where \(x = 0\).

1. (a) By writing \(\operatorname { cosec } x\) as \(\frac { 1 } { \sin x }\), show that

$$\frac { \mathrm { d } ( \operatorname { cosec } x ) } { \mathrm { d } x } = - \operatorname { cosec } x \cot x$$ Given that \(y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }\),
(b) find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). The curve with equation \(y = \mathrm { e } ^ { 3 x } \operatorname { cosec } 2 x , 0 < x < \frac { \pi } { 2 }\), has a single turning point.
(c) Show that the \(x\)-coordinate of this turning point is at \(x = \frac { 1 } { 2 } \arctan k\) where the value
of the constant \(k\) should be found. of the constant \(k\) should be found.

6. A curve \(C\) has equation $$y = x ^ { \sin x } \quad x > 0 \quad y > 0$$

1. Find, by firstly taking natural logarithms, an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Hence show that the \(x\) coordinates of the stationary points of \(C\) are solutions of the equation $$\tan x + x \ln x = 0$$

The equation of a curve is \(y = \frac{1+x}{1+2x}\) for \(x > -\frac{1}{2}\). Show that the gradient of the curve is always negative. [3]

The function \(\text{f}(x) = \frac{\sin x}{2 - \cos x}\) has domain \(-\pi \leqslant x \leqslant \pi\). Fig. 8 shows the graph of \(y = \text{f}(x)\) for \(0 \leqslant x \leqslant \pi\). \includegraphics{figure_6}

1. Find \(\text{f}(-x)\) in terms of \(\text{f}(x)\). Hence sketch the graph of \(y = \text{f}(x)\) for the complete domain \(-\pi \leqslant x \leqslant \pi\). [3]
2. Show that \(\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}\). Hence find the exact coordinates of the turning point P. State the range of the function \(\text{f}(x)\), giving your answer exactly. [8]
3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int_0^\pi \frac{\sin x}{2 - \cos x} dx\). [4]
4. Sketch the graph of \(y = \text{f}(2x)\). [1]
5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx\). [2]