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

\includegraphics{figure_11} The diagram shows part of the curve \(y = (x + 1)^2 + (x + 1)^{-1}\) and the line \(x = 1\). The point \(A\) is the minimum point on the curve.

1. Show that the \(x\)-coordinate of \(A\) satisfies the equation \(2(x + 1)^3 = 1\) and find the exact value of \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) at \(A\). [5]
2. Find, showing all necessary working, the volume obtained when the shaded region is rotated through \(360°\) about the \(x\)-axis. [6]

Find the exact value of \(\int_1^4 \frac{\ln x}{\sqrt{x}} dx\). [5]

The equation of a curve is \(y = x \ln(8 - x)\). The gradient of the curve is equal to 1 at only one point, when \(x = a\).

1. Show that \(a\) satisfies the equation \(x = 8 - \frac{8}{\ln(8 - x)}\). [3]
2. Verify by calculation that \(a\) lies between 2.9 and 3.1. [2]
3. Use an iterative formula based on the equation in part (i) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

It is given that \(y = \ln(ax + 1)\), where \(a\) is a positive constant. Prove by mathematical induction that, for every positive integer \(n\), $$\frac{d^n y}{dx^n} = (-1)^{n-1} \frac{(n-1)!a^n}{(ax+1)^n}.$$ [6]

It is given that \(y = \ln(ax + 1)\), where \(a\) is a positive constant. Prove by mathematical induction that, for every positive integer \(n\), $$\frac{d^n y}{dx^n} = (-1)^{n-1} \frac{(n-1)! a^n}{(ax+1)^n}.$$ [6]

The curve \(C\) with equation \(y = f(x)\), \(x \neq 0\), passes through the point \((3, 7\frac{1}{2})\). Given that \(f'(x) = 2x + \frac{3}{x^2}\),

1. find \(f(x)\). [5]
2. Verify that \(f(-2) = 5\). [1]
3. Find an equation for the tangent to \(C\) at the point \((-2, 5)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [4]

The curve \(C\) has equation \(y = f(x)\), \(x \neq 0\), and the point \(P(2, 1)\) lies on \(C\). Given that $$f'(x) = 3x^2 - 6 - \frac{8}{x^3},$$

1. find \(f(x)\). [5]
2. Find an equation for the tangent to \(C\) at the point \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are integers. [4]

\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation $$y = 9 - 2x - \frac{2}{\sqrt{x}}, \quad x > 0.$$ The point \(A(1, 5)\) lies on \(C\) and the curve crosses the \(x\)-axis at \(B(b, 0)\), where \(b\) is a constant and \(b > 0\).

1. Verify that \(b = 4\). [1]

The tangent to \(C\) at the point \(A\) cuts the \(x\)-axis at the point \(D\), as shown in Fig. 2.

1. Show that an equation of the tangent to \(C\) at \(A\) is \(y + x = 6\). [4]
2. Find the coordinates of the point \(D\). [1]

The shaded region \(R\), shown in Fig. 2, is bounded by \(C\), the line \(AD\) and the \(x\)-axis.

1. Use integration to find the area of \(R\). [6]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = f(x)\), where $$f(x) = 10 + \ln(3x) - \frac{1}{2}e^x, \quad 0.1 \leq x \leq 3.3.$$ Given that \(f(k) = 0\),

1. show, by calculation, that \(3.1 < k < 3.2\). [2]
2. Find \(f'(x)\). [3]

The tangent to the graph at \(x = 1\) intersects the \(y\)-axis at the point \(P\).

1. 1. Find an equation of this tangent.
   2. Find the exact \(y\)-coordinate of \(P\), giving your answer in the form \(a + \ln b\). [5]

The curve with equation \(y = \ln 3x\) crosses the \(x\)-axis at the point \(P (p, 0)\).

1. Sketch the graph of \(y = \ln 3x\), showing the exact value of \(p\). [2]

The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.

1. Show that \(x = q\) is a solution of the equation \(x^2 + \ln 3x = 0\). [4]
2. Show that the equation in part (b) can be rearranged in the form \(x = \frac{1}{3}e^{-x^2}\). [2]
3. Use the iteration formula \(x_{n+1} = \frac{1}{3}e^{-x_n^2}\), with \(x_0 = \frac{1}{4}\), to find \(x_1, x_2, x_3\) and \(x_4\). Hence write down, to 3 decimal places, an approximation for \(q\). [3]

The curve \(C\) has equation \(y = f(x)\), where $$f(x) = 3 \ln x + \frac{1}{x}, \quad x > 0.$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\). [4]
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. [2]

The point \(Q\) on \(C\) has \(x\)-coordinate \(1\).

1. Find an equation for the normal to \(C\) at \(Q\). [4]

The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).

1. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac{2}{x} - 3 = 0\),
   2. lies between \(0.13\) and \(0.14\). [4]

Given that \(y = \log_a x, x > 0\), where \(a\) is a positive constant,

1. 1. express \(x\) in terms of \(a\) and \(y\), [1]
   2. deduce that \(\ln x = y \ln a\). [1]
2. Show that \(\frac{dy}{dx} = \frac{1}{x \ln a}\). [2]

The curve \(C\) has equation \(y = \log_{10} x, x > 0\). The point \(A\) on \(C\) has \(x\)-coordinate \(10\). Using the result in part (b),

1. find an equation for the tangent to \(C\) at \(A\). [4]

The tangent to \(C\) at \(A\) crosses the \(x\)-axis at the point \(B\).

1. Find the exact \(x\)-coordinate of \(B\). [2]

The diagram shows the curve with equation \(y = \ln(6x)\). \includegraphics{figure_1}

1. State the \(x\)-coordinate of the point of intersection of the curve with the \(x\)-axis. [1]
2. Find \(\frac{dy}{dx}\). [2]
3. Use Simpson's rule with 6 strips (7 ordinates) to find an estimate for \(\int_1^7 \ln(6x) \, dx\), giving your answer to three significant figures. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 10 + \ln(3x) - \frac{1}{2}e^x, \quad 0.1 \leq x \leq 3.3.$$ Given that f(k) = 0,

1. show, by calculation, that \(3.1 < k < 3.2\). [2]
2. Find f'(x). [3]

The tangent to the graph at \(x = 1\) intersects the \(y\)-axis at the point \(P\).

1. 1. Find an equation of this tangent.
   2. Find the exact \(y\)-coordinate of \(P\), giving your answer in the form \(a + \ln b\). [5]

The curve \(C\) has equation \(y = \text{f}(x)\), where $$\text{f}(x) = 3 \ln x + \frac{1}{x}, \quad x > 0.$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\). [4]
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. [2]

The point \(Q\) on \(C\) has \(x\)-coordinate 1.

1. Find an equation for the normal to \(C\) at \(Q\). [4]

The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).

1. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac{2}{x} - 3 = 0\),
   2. lies between 0.13 and 0.14. [4]

Given that \(y = \log_a x\), \(x > 0\), where \(a\) is a positive constant,

1. 1. express \(x\) in terms of \(a\) and \(y\), [1]
   2. deduce that \(\ln x = y \ln a\). [1]
2. Show that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x \ln a}\). [2]

The curve \(C\) has equation \(y = \log_{10} x\), \(x > 0\). The point \(A\) on \(C\) has \(x\)-coordinate 10. Using the result in part (b),

1. find an equation for the tangent to \(C\) at \(A\). [4]

The tangent to \(C\) at \(A\) crosses the \(x\)-axis at the point \(B\).

1. Find the exact \(x\)-coordinate of \(B\). [2]

The curve with equation \(y = \ln 3x\) crosses the \(x\)-axis at the point \(P(p, 0)\).

1. Sketch the graph of \(y = \ln 3x\), showing the exact value of \(p\). [2]

The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.

1. Show that \(x = q\) is a solution of the equation \(x^2 + \ln 3x = 0\). [4]
2. Show that the equation in part (b) can be rearranged in the form \(x = \frac{1}{3}e^{-x^2}\). [2]
3. Use the iteration formula \(x_{n + 1} = \frac{1}{3}e^{-x_n^2}\), with \(x_0 = \frac{1}{3}\), to find \(x_1, x_2, x_3\) and \(x_4\). Hence write down, to 3 decimal places, an approximation for \(q\). [3]

\includegraphics{figure_8} The diagram shows part of the curve \(y = \ln(5 - x^2)\) which meets the \(x\)-axis at the point \(P\) with coordinates \((2, 0)\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(PQ\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\). [5]
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures. [4]
3. Deduce an approximation to the area of the region \(B\). [2]

It is given that \(y = 5^{x-1}\).

1. Show that \(x = 1 + \frac{\ln y}{\ln 5}\). [2]
2. Find an expression for \(\frac{dx}{dy}\) in terms of \(y\). [2]
3. Hence find the exact value of the gradient of the curve \(y = 5^{x-1}\) at the point \((3, 25)\). [2]

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

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

The function \(f(x) = \ln(1 + x^2)\) has domain \(-3 \leq x \leq 3\). Fig. 9 shows the graph of \(y = f(x)\). \includegraphics{figure_9}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve. [3]
2. Find the gradient of the curve at the point P\((2, \ln 5)\). [4]
3. Explain why the function does not have an inverse for the domain \(-3 \leq x \leq 3\). [1]

The domain of \(f(x)\) is now restricted to \(0 \leq x \leq 3\). The inverse of \(f(x)\) is the function \(g(x)\).

1. Sketch the curves \(y = f(x)\) and \(y = g(x)\) on the same axes. State the domain of the function \(g(x)\). Show that \(g(x) = \sqrt{e^x - 1}\). [6]
2. Differentiate \(g(x)\). Hence verify that \(g'(\ln 5) = \frac{1}{4}\). Explain the connection between this result and your answer to part (ii). [5]

Fig. 9 shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The function \(y = \text{f}(x)\) is given by $$\text{f}(x) = \ln \left( \frac{2x}{1+x} \right), \quad x > 0.$$ The curve \(y = \text{f}(x)\) crosses the \(x\)-axis at P, and the line \(x = 2\) at Q. \includegraphics{figure_9}

1. Verify that the \(x\)-coordinate of P is 1. Find the exact \(y\)-coordinate of Q. [2]
2. Find the gradient of the curve at P. [Hint: use \(\frac{a}{b} = \ln a - \ln b\).] [4]

The function \(\text{g}(x)\) is given by $$\text{g}(x) = \frac{e^x}{2-e^x}, \quad x < \ln 2.$$ The curve \(y = \text{g}(x)\) crosses the \(y\)-axis at the point R.

1. Show that \(\text{g}(x)\) is the inverse function of \(\text{f}(x)\). Write down the gradient of \(y = \text{g}(x)\) at R. [5]
2. Show, using the substitution \(u = 2 - e^x\) or otherwise, that \(\int_0^{\ln \frac{4}{3}} \text{g}(x) dx = \ln \frac{3}{2}\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac{32}{27}\). [Hint: consider its reflection in \(y = x\).] [7]

The volume \(V\) m³ of a pile of grain of height \(h\) metres is modelled by the equation $$V = 4\sqrt{h^3 + 1} - 4.$$

1. Find \(\frac{dV}{dh}\) when \(h = 2\). [4]

At a certain time, the height of the pile is 2 metres, and grain is being added so that the volume is increasing at a rate of 0.4 m³ per minute.

1. Find the rate at which the height is increasing at this time. [3]