1.07l Derivative of ln(x): and related functions

The function f(x) is defined as \(f(x) = \frac{\ln x}{x}\). The graph of the function is shown in Fig. 6. \includegraphics{figure_6}

1. Give the coordinates of the point, P, where the curve crosses the \(x\)-axis. [1]
2. Use calculus to find the coordinates of the stationary point, Q, and show that it is a maximum. [6]

$$f(x) = 2x^2 + 3 \ln (2 - x), \quad x \in \mathbb{R}, \quad x < 2.$$

1. Show that the equation \(f(x) = 0\) can be written in the form $$x = 2 - e^{kx^2},$$ where \(k\) is a constant to be found. [3]

The root, \(\alpha\), of the equation \(f(x) = 0\) is \(1.9\) correct to \(1\) decimal place.

1. Use the iteration formula $$x_{n+1} = 2 - e^{kx_n^2},$$ with \(x_0 = 1.9\) and your value of \(k\), to find \(\alpha\) to \(3\) decimal places and justify the accuracy of your answer. [5]
2. Solve the equation \(f'(x) = 0\). [5]

Show that the curve \(y = x^2 \ln x\) has a stationary point when \(x = \frac{1}{\sqrt{e}}\). [6]

A curve has equation \(y = \frac{x}{2 + 3\ln x}\). Find \(\frac{dy}{dx}\). Hence find the exact coordinates of the stationary point of the curve. [7]

A curve is defined by the equation \(y = 2x \ln(1 + x)\).

1. Find \(\frac{dy}{dx}\) and hence verify that the origin is a stationary point of the curve. [4]
2. Find \(\frac{d^2y}{dx^2}\) and use this to verify that the origin is a minimum point. [5]
3. Using the substitution \(u = 1 + x\), show that \(\int \frac{x^2}{1+x} \, dx = \int \left(u - 2 + \frac{1}{u}\right) du\). Hence evaluate \(\int_0^1 \frac{x^2}{1+x} \, dx\), giving your answer in an exact form. [6]
4. Using integration by parts and your answer to part (iii), evaluate \(\int_0^1 2x \ln(1 + x) \, dx\). [4]

Differentiate each of the following with respect to \(x\) and simplify your answers.

1. \(\ln(\cos x)\) [2]
2. \(x^2 \sin 3x\) [2]

It is given that \(f(x) = x^2 - \sin x\).

1. The iteration \(x_{n+1} = \sqrt{\sin x_n}\), with \(x_1 = 0.875\), is to be used to find a real root, \(\alpha\), of the equation \(f(x) = 0\). Find \(x_2, x_3\) and \(x_4\), giving the answers correct to 6 decimal places. [2]
2. The error \(e_n\) is defined by \(e_n = \alpha - x_n\). Given that \(\alpha = 0.876726\), correct to 6 decimal places, find \(e_3\) and \(e_4\). Given that \(g(x) = \sqrt{\sin x}\), use \(e_3\) and \(e_4\) to estimate \(g'(\alpha)\). [3]

It is given that \(f(x) = \tan^{-1}(1 + x)\).

1. Find \(f(0)\) and \(f'(0)\), and show that \(f''(0) = -\frac{1}{2}\). [4]
2. Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^2\). [2]

Given that \(f(x) = \ln(\cos 3x)\), find \(f'(0)\) and \(f''(0)\). Hence show that the first term in the Maclaurin series for \(f(x)\) is \(ax^2\), where the value of \(a\) is to be found. [4]

1. Prove that the derivative of \(\cos^{-1} x\) is \(-\frac{1}{\sqrt{1 - x^2}}\). [3]

A curve has equation \(y = \cos^{-1}(1 - x^2)\), for \(0 < x < \sqrt{2}\).

1. Find and simplify \(\frac{dy}{dx}\), and hence show that $$(2 - x^2)\frac{d^2y}{dx^2} = x\frac{dy}{dx}.$$ [5]

1. Given that \(y = \ln [t + \sqrt{(1 + t^2)}]\), show that \(\frac{dy}{dt} = \frac{1}{\sqrt{(1+t^2)}}\). [3]

The curve \(C\) has parametric equations $$x = \frac{1}{\sqrt{(1+t^2)}}, \quad y = \ln [t + \sqrt{(1 + t^2)}], \quad t \in \mathbb{R}.$$ A student was asked to prove that, for \(t > 0\), the gradient of the tangent to \(C\) is negative. The attempted proof was as follows: $$y = \ln \left(t + \frac{1}{x}\right)$$ $$= \ln \left(\frac{tx + 1}{x}\right)$$ $$= \ln (tx + 1) - \ln x$$ $$\therefore \frac{dy}{dx} = \frac{t}{tx + 1} - \frac{1}{x}$$ $$= \frac{\frac{t}{x}}{t + \frac{1}{x}} - \frac{1}{x}$$ $$= \frac{t\sqrt{(1+t^2)}}{t + \sqrt{(1+t^2)}} - \sqrt{(1 + t^2)}$$ $$= -\frac{(1+t^2)}{t + \sqrt{(1+t^2)}}$$ As \((1 + t^2) > 0\), and \(t + \sqrt{(1 + t^2)} > 0\) for \(t > 0\), \(\frac{dy}{dx} < 0\) for \(t > 0\).

1. 1. Identify the error in this attempt.
   2. Give a correct version of the proof. [6]
2. Prove that \(\ln [-t + \sqrt{(1 + t^2)}] = -\ln [t + \sqrt{(1 + t^2)}]\). [3]
3. Deduce that \(C\) is symmetric about the \(x\)-axis and sketch the graph of \(C\). [3]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve \(C\) with equation $$y = \cos x \ln(\sec x), \quad -\frac{\pi}{2} < x < \frac{\pi}{2}$$ The points \(A\) and \(B\) are maximum points on \(C\).

1. Find the coordinates of \(B\) in terms of e. [5]

The finite region \(R\) lies between \(C\) and the line \(AB\).

1. Show that the area of \(R\) is $$\frac{2}{e} \arccos \left(\frac{1}{e}\right) + 2\ln \left(e + \sqrt{(e^2 - 1)}\right) - \frac{4}{e} \sqrt{(e^2 - 1)}.$$ [arccos \(x\) is an alternative notation for \(\cos^{-1} x\)] [8]

In this question you must show detailed reasoning. \includegraphics{figure_5} The diagram shows the curve with equation \(y = \frac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$4x^3 + 3x - 3 = 0.$$ [5]
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]
3. Show that the iteration $$x_{n+1} = \frac{3 - 4x_n^3}{3}$$ cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]
4. Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]

At the point \((1, 0)\) on the curve \(y = \ln x\), which statement below is correct? Tick (\(\checkmark\)) one box. [1 mark] The gradient is negative and decreasing The gradient is negative and increasing The gradient is positive and decreasing The gradient is positive and increasing

Given that \(\frac{dy}{dx} = \frac{1}{x}\) find \(\frac{d^2y}{dx^2}\) Circle your answer. [1 mark] \(-\frac{2}{x^2}\) \(-\frac{1}{x^2}\) \(\frac{1}{x^2}\) \(\frac{2}{x^2}\)

\includegraphics{figure_3} The curve \(C_1\), shown in Figure 3, has equation \(y = 4x^2 - 6x + 4\). The point \(P\left(\frac{1}{2}, 2\right)\) lies on \(C_1\) The curve \(C_2\), also shown in Figure 3, has equation \(y = \frac{1}{2}x + \ln(2x)\). The normal to \(C_1\) at the point \(P\) meets \(C_2\) at the point \(Q\). Find the exact coordinates of \(Q\). (Solutions based entirely on graphical or numerical methods are not acceptable.) [8]

Find the gradient of the tangent to the curve $$y = \sin^{-1} x$$ at the point where \(x = \frac{1}{5}\) Circle your answer. [1 mark] \(\frac{5\sqrt{6}}{12}\) \quad \(\frac{2\sqrt{6}}{5}\) \quad \(\frac{4\sqrt{3}}{25}\) \quad \(\frac{25}{24}\)

You are given that \(\mathrm{f}(x) = \tan^{-1}(1 + x)\).

1. 1. Find the value of \(\mathrm{f}(0)\). [1]
   2. Determine the value of \(\mathrm{f}'(0)\). [2]
   3. Show that \(\mathrm{f}''(0) = -\frac{1}{2}\). [3]
2. Hence find the Maclaurin series for \(\mathrm{f}(x)\) up to and including the term in \(x^2\). [2]

Differentiate the following with respect to \(x\), simplifying your answers fully

1. \(y = e^{3x} + \ln 2x\) [1]
2. \(y = (5 + x^2)^{\frac{3}{2}}\) [2]
3. \(y = \frac{2x}{(5-3x^2)^{\frac{1}{2}}}\) [4]
4. \(y = e^{-\frac{3}{x}} \ln(1 + x^3)\) [3]