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

7

1. Given that \(f ( x ) = x + \frac { 3 } { x }\), find \(f ^ { \prime } ( x )\).
2. Find the gradient of the curve \(\mathrm { y } = \mathrm { x } ^ { \frac { 5 } { 2 } }\) at the point where \(\mathrm { x } = 4\).

9. A curve has the equation \(y = \frac { x } { 2 } + 3 - \frac { 1 } { x } , x \neq 0\). The point \(A\) on the curve has \(x\)-coordinate 2 .

1. Find the gradient of the curve at \(A\).
2. Show that the tangent to the curve at \(A\) has equation $$3 x - 4 y + 8 = 0$$ The tangent to the curve at the point \(B\) is parallel to the tangent at \(A\).
3. Find the coordinates of \(B\).

9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

9. The diagram shows the curve with equation \(y = 2 x - 3 \ln ( 2 x + 5 )\) and the normal to the curve at the point \(P ( - 2 , - 4 )\).

1. Find an equation for the normal to the curve at \(P\). The normal to the curve at \(P\) intersects the curve again at the point \(Q\) with \(x\)-coordinate \(q\).
2. Show that \(1 < q < 2\).
3. Show that \(q\) is a solution of the equation $$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
4. Use an iterative process based on the equation above with a starting value of 1.5 to find the value of \(q\) to 3 significant figures and justify the accuracy of your answer.

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

1. \(\quad \ln ( 3 x - 2 )\)
2. \(\frac { 2 x + 1 } { 1 - x }\)
3. \(x ^ { \frac { 3 } { 2 } } \mathrm { e } ^ { 2 x }\)

3. The curve \(C\) has the equation \(y = 2 \mathrm { e } ^ { x } - 6 \ln x\) and passes through the point \(P\) with \(x\)-coordinate 1.

1. Find an equation for the tangent to \(C\) at \(P\). The tangent to \(C\) at \(P\) meets the coordinate axes at the points \(Q\) and \(R\).
2. Show that the area of triangle \(O Q R\), where \(O\) is the origin, is \(\frac { 9 } { 3 - \mathrm { e } }\).

A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).

1. Show that the tangent to the curve at the point where \(x = 1\) has the equation $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
2. Show that \(0.3 < \alpha < 0.4\)
3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(\alpha\) correct to 5 decimal places.
   You should show the result of each iteration.

5. A curve has the equation \(y = \sqrt { 3 x + 11 }\). The point \(P\) on the curve has \(x\)-coordinate 3 .

1. Show that the tangent to the curve at \(P\) has the equation $$3 x - 4 \sqrt { 5 } y + 31 = 0$$ The normal to the curve at \(P\) crosses the \(y\)-axis at \(Q\).
2. Find the \(y\)-coordinate of \(Q\) in the form \(k \sqrt { 5 }\).

7. The curve with equation \(y = x ^ { \frac { 5 } { 2 } } \ln \frac { x } { 4 } , x > 0\) crosses the \(x\)-axis at the point \(P\).

1. Write down the coordinates of \(P\). The normal to the curve at \(P\) crosses the \(y\)-axis at the point \(Q\).
2. Find the area of triangle \(O P Q\) where \(O\) is the origin. The curve has a stationary point at \(R\).
3. Find the \(x\)-coordinate of \(R\) in exact form.

8 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-4_787_742_276_719} The diagram shows part of the curve \(y = \ln \left( 5 - x ^ { 2 } \right)\) 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 \(P Q\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\).
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.
3. Deduce an approximation to the area of the region \(B\).

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

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

8

1. Given that \(\mathrm { y } = \frac { 4 \ln \mathrm { x } - 3 } { 4 \ln \mathrm { x } + 3 }\), show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = \frac { 24 } { \mathrm { x } ( 4 \ln \mathrm { x } + 3 ) ^ { 2 } }\).
2. Find the exact value of the gradient of the curve \(y = \frac { 4 \ln x - 3 } { 4 \ln x + 3 }\) at the point where it crosses the \(x\)-axis.
3. \includegraphics[max width=\textwidth, alt={}, center]{133c38fb-307f-4f20-86cb-1bd57cc4f870-3_524_830_941_699} The diagram shows part of the curve with equation $$\mathrm { y } = \frac { 2 } { \mathrm { x } ^ { \frac { 1 } { 2 } } ( 4 \ln \mathrm { x } + 3 ) }$$ The region shaded in the diagram is bounded by the curve and the lines \(x = 1 , x = e\) and \(y = 0\). Find the exact volume of the solid produced when this shaded region is rotated completely about the x -axis.

3 Find, in the form \(y = m x + c\), the equation of the tangent to the curve $$y = x ^ { 2 } \ln x$$ at the point with \(x\)-coordinate e.

8 \includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_476_608_287_756} The diagram shows the curve \(y = ( \ln x ) ^ { 2 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. The point \(P\) on the curve is the point at which the gradient takes its maximum value. Show that the tangent at \(P\) passes through the point \(( 0 , - 1 )\).

7 Fig. 7 shows the curve $$y = 2 x - x \ln x , \text { where } x > 0 .$$ The curve crosses the \(x\)-axis at A , and has a turning point at B . The point C on the curve has \(x\)-coordinate 1 . Lines CD and BE are drawn parallel to the \(y\)-axis. \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of A , giving your answer in terms of e .
2. Find the exact coordinates of B .
3. Show that the tangents at A and C are perpendicular to each other.
4. Using integration by parts, show that $$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$ Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines CD and BE . \section*{[Question 8 is printed overleaf.]}

1

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

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

8 Fig. 8 shows the curve \(y = 3 \ln x + x - x ^ { 2 }\).
The curve crosses the \(x\)-axis at P and Q , and has a turning point at R . The \(x\)-coordinate of Q is approximately 2.05 . \begin{figure}[h] \end{figure}

1. Verify that the coordinates of P are \(( 1,0 )\).
2. Find the coordinates of R , giving the \(y\)-coordinate correct to 3 significant figures. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that R is a maximum point.
3. Find \(\int \ln x \mathrm {~d} x\). Hence calculate the area of the region enclosed by the curve and the \(x\)-axis between P and Q , giving your answer to 2 significant figures.

6 Find \(\frac { \mathrm { d } } { \mathrm { d } x } ( x \ln x )\) and hence or otherwise find the value of \(\int _ { 2 } ^ { 3 } \ln x \mathrm {~d} x\), giving your answer in the form \(\ln a + b\), where \(a\) and \(b\) are to be determined.

8 You are given that \(\mathrm { f } ( x ) = \frac { x } { x ^ { 2 } + 1 }\) for all real values of \(x\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\).
2. Hence show that there is a stationary value at \(\left( 1 , \frac { 1 } { 2 } \right)\) and find the coordinates of the other stationary point.
3. The graph of the curve is shown in Fig. 8. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-3_518_892_1612_705} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure} State whether the curve is odd or even and prove the result algebraically.
4. Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x = \int _ { a } ^ { b } k \frac { 1 } { u + 1 } \mathrm {~d} u\), where the values of \(a , b\) and \(k\) are to be determined.
5. Hence find the area of the shaded region in Fig. 8.

3 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). 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 \(y = x\).]

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

3 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

3 The function \(f ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h] \end{figure} (1) Show algebraically that the function is even. State how this property relates to the shape of the curve.
(ii) Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
(iii) Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\),
(iv) Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(g ( x )\).
Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
(v) Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the commection between this result and your answer to part (ii).