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

11 The point \(P\) on the curve \(y = k \sqrt { x }\) has \(x\)-coordinate 4 . The normal to the curve at \(P\) is parallel to the line \(2 x + 3 y = 0\).

1. Find the value of \(k\).
2. This normal meets the \(x\)-axis at the point \(Q\). Calculate the area of the triangle \(O P Q\), where \(O\) is the point \(( 0,0 )\). RECOGNISING ACHIEVEMENT

6 Find the gradient of the curve \(y = 2 x + \frac { 6 } { \sqrt { x } }\) at the point where \(x = 4\).

1 Given that \(y = 4 x ^ { 2 } \ln x\), find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) when \(x = \mathrm { e } ^ { 2 }\).

6 The curves \(C _ { 1 }\) and \(C _ { 2 }\) have equations $$y = \ln ( 4 x - 7 ) + 18 \quad \text { and } \quad y = a \left( x ^ { 2 } + b \right) ^ { \frac { 1 } { 2 } }$$ respectively, where \(a\) and \(b\) are positive constants. The point \(P\) lies on both curves and has \(x\)-coordinate 2 . It is given that the gradient of \(C _ { 1 }\) at \(P\) is equal to the gradient of \(C _ { 2 }\) at \(P\). Find the values of \(a\) and \(b\).

8 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 \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

6 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \leqslant x \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\). The point Q is the reflection of P in \(y = x\).
2. Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q . Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .

5 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 Find the exact value of \(\int _ { 1 } ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x\).

7

1. Given that \(y = \ln ( 1 + \sin x ) - \ln ( \cos x )\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \cos x }\).
2. Using this result, evaluate \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \sec x \mathrm {~d} x\), giving your answer as a single logarithm.

6

1. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
2. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).

2 It is given that \(\mathrm { f } ( x ) = \tanh ^ { - 1 } x\).

1. Show that \(\mathrm { f } ^ { \prime \prime \prime } ( x ) = \frac { 2 \left( 1 + 3 x ^ { 2 } \right) } { \left( 1 - x ^ { 2 } \right) ^ { 3 } }\).
2. Hence find the Maclaurin series for \(\mathrm { f } ( x )\), up to and including the term in \(x ^ { 3 }\).

1 It is given that \(\mathrm { f } ( x ) = \tan ^ { - 1 } 2 x\) and \(\mathrm { g } ( x ) = p \tan ^ { - 1 } x\), where \(p\) is a constant. Find the value of \(p\) for which \(\mathrm { f } ^ { \prime } \left( \frac { 1 } { 2 } \right) = \mathrm { g } ^ { \prime } \left( \frac { 1 } { 2 } \right)\).

4 The curves \(y = \cos ^ { - 1 } x\) and \(y = \tan ^ { - 1 } ( \sqrt { 2 } x )\) intersect at a point \(A\).

1. Verify that the coordinates of \(A\) are \(\left( \frac { 1 } { \sqrt { 2 } } , \frac { 1 } { 4 } \pi \right)\).
2. Determine whether the tangents to the curves at \(A\) are perpendicular.

2 Prove by mathematical induction that, for all positive integers \(n\), $$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \frac { 1 } { 2 x + 3 } \right) = ( - 1 ) ^ { n } \frac { n ! 2 ^ { n } } { ( 2 x + 3 ) ^ { n + 1 } }$$

2

1. Given that \(y = \ln ( 4 + 3 x )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. Hence, by using Maclaurin's theorem, find the first three terms in the expansion, in ascending powers of \(x\), of \(\ln ( 4 + 3 x )\).
3. Write down the first three terms in the expansion, in ascending powers of \(x\), of \(\ln ( 4 - 3 x )\).
4. Show that, for small values of \(x\), $$\ln \left( \frac { 4 + 3 x } { 4 - 3 x } \right) \approx \frac { 3 } { 2 } x$$

6

1. Given that \(y = \ln \cos 2 x\), find \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\).
2. Use Maclaurin's theorem to show that the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \cos 2 x\) are \(- 2 x ^ { 2 } - \frac { 4 } { 3 } x ^ { 4 }\).
3. Hence find the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \sec ^ { 2 } 2 x\).

6

1. It is given that \(y = \ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\).
   1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \tan x\).
   2. Find \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\).
2. Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( \mathrm { e } ^ { 3 x } \cos x \right)\) are \(3 x - \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 12 } x ^ { 4 }\).
   (3 marks)
3. Write down the expansion of \(\ln ( 1 + p x )\), where \(p\) is a constant, in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\).
4. 1. Find the value of \(p\) for which \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) exists.
   2. Hence find the value of \(\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]\) when \(p\) takes the value found in part (d)(i).

6 The function f is defined by $$\mathrm { f } ( x ) = ( 9 + \tan x ) ^ { \frac { 1 } { 2 } }$$

1. 1. Find \(f ^ { \prime \prime } ( x )\).
   2. By using Maclaurin's theorem, show that, for small values of \(x\), $$( 9 + \tan x ) ^ { \frac { 1 } { 2 } } \approx 3 + \frac { x } { 6 } - \frac { x ^ { 2 } } { 216 }$$
2. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { f ( x ) - 3 } { \sin 3 x } \right]$$

5

1. Write down the expansion of \(\cos 4 x\) in ascending powers of \(x\) up to and including the term in \(x ^ { 4 }\). Give your answer in its simplest form.
2. 1. Given that \(y = \ln \left( 2 - \mathrm { e } ^ { x } \right)\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) and \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\).
      (You may leave your expression for \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) unsimplified.)
   2. Hence, by using Maclaurin's theorem, show that the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln \left( 2 - \mathrm { e } ^ { x } \right)\) are $$- x - x ^ { 2 } - x ^ { 3 }$$
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { x \ln \left( 2 - \mathrm { e } ^ { x } \right) } { 1 - \cos 4 x } \right]$$

5

1. Given that \(y = \ln ( 1 + 2 \tan x )\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
   (You may leave your expression for \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) unsimplified.)
2. Hence, using Maclaurin's theorem, find the first two non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln ( 1 + 2 \tan x )\).
   (2 marks)
3. Find $$\lim _ { x \rightarrow 0 } \left[ \frac { \ln ( 1 + 2 \tan x ) } { \ln ( 1 - x ) } \right]$$ (4 marks)

6 It is given that \(y = \ln ( 1 + \sin x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - \mathrm { e } ^ { - y }\).
3. Express \(\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }\) in terms of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\mathrm { e } ^ { - y }\).
4. Hence, by using Maclaurin's theorem, find the first four non-zero terms in the expansion, in ascending powers of \(x\), of \(\ln ( 1 + \sin x )\).

5

1. Differentiate \(\ln ( \ln x )\) with respect to \(x\).
2. 1. Show that \(\ln x\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 1 } { x \ln x } y = 9 x ^ { 2 } , \quad x > 1$$
   2. Hence find the solution of this differential equation, given that \(y = 4 \mathrm { e } ^ { 3 }\) when \(x = \mathrm { e }\).
      (6 marks)

11 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } \ln x } { \mathrm { e } ^ { 3 y } }\).
The curve passes through the point (e, 1).

1. Find the equation of this curve, giving your answer in the form \(\mathrm { e } ^ { 3 y } = \mathrm { f } ( x )\).
2. Show that, when \(x = \mathrm { e } ^ { 2 }\), the \(y\)-coordinate of this curve can be written as \(y = a + \frac { 1 } { 3 } \ln \left( b \mathrm { e } ^ { 3 } + c \right)\), where \(a , b\) and \(c\) are constants to be determined.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where \(x \in R\), \(x > 0\) $$\mathrm { f } ( x ) = ( 0.5 x - 8 ) \ln ( x + 1 ) \quad 0 \leq x \leq A$$ a. Find the value of \(A\).
b. Find \(\mathrm { f } ^ { \prime } ( x )\) The curve has a minimum turning point at \(B\).
c. Show that the \(x\)-coordinate of \(B\) is a solution of the equation $$x = \frac { 17 } { \ln ( x + 1 ) + 1 } - 1$$ d. Use the iteration formula $$x _ { n + 1 } = \frac { 17 } { \ln \left( x _ { n } + 1 \right) + 1 } - 1$$ with \(x _ { 0 } = 5\) to find the values of \(x _ { 1 }\) and the value of \(x _ { 6 }\) giving your answers to three decimal places.

6. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \frac { 2 x ^ { 2 } - x } { \sqrt { x } } - 2 \ln \left( \frac { x } { 2 } \right) , \quad x > 0$$ The curve has a minimum turning point at \(Q\), as shown in Figure 4.
a. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x ^ { 2 } - x - 4 \sqrt { x } } { 2 x \sqrt { x } }\) b. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \sqrt { \frac { x } { 6 } + \frac { 2 \sqrt { x } } { 3 } }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \sqrt { \frac { x _ { n } } { 6 } + \frac { 2 \sqrt { x _ { n } } } { 3 } }$$ is used.
c. Taking \(x _ { 0 } = 0.8\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.