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

4 The parametric equations of a curve are $$x = 2 t - \tan t , \quad y = \ln ( \sin 2 t )$$ for \(0 < t < \frac { 1 } { 2 } \pi\).
Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \cot t\).

2 It is given that \(y = 2 ^ { x }\).

1. By differentiating \(\ln y\) with respect to \(x\), show that \(\frac { \mathrm { dy } } { \mathrm { dx } } = 2 ^ { \mathrm { x } } \ln 2\).
2. Write down \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\).
3. Hence find the first three terms in the Maclaurin's series for \(2 ^ { X }\).

1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where \(x > 0\)

Given that

• \(f ^ { \prime } ( x ) = \frac { 4 x ^ { 2 } + 10 - 7 x ^ { \frac { 1 } { 2 } } } { 4 x ^ { \frac { 1 } { 2 } } }\)
• the point \(P ( 4 , - 1 )\) lies on \(C\)
  1. (i) find the value of the gradient of \(C\) at \(P\) (ii) Hence find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers to be found.
  2. Find \(\mathrm { f } ( x )\).

2. $$\mathrm { f } ( x ) = \frac { 8 } { x ^ { 2 } } - 4 \sqrt { x } + 3 x - 1 , \quad x > 0$$ Giving your answers in their simplest form, find

1. \(\mathrm { f } ^ { \prime } ( x )\)
2. \(\int \mathrm { f } ( x ) \mathrm { d } x\)

10. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve \(C\) with equation $$y = 2 - \frac { 1 } { x } , \quad x \neq 0$$ The curve crosses the \(x\)-axis at the point \(A\).

1. Find the coordinates of \(A\).
2. Show that the equation of the normal to \(C\) at \(A\) can be written as $$2 x + 8 y - 1 = 0$$ The normal to \(C\) at \(A\) meets \(C\) again at the point \(B\), as shown in Figure 2 .
3. Find the coordinates of \(B\).

10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 1 } { 2 } x + \frac { 27 } { x } - 12 , \quad x > 0$$ The point \(A\) lies on \(C\) and has coordinates \(\left( 3 , - \frac { 3 } { 2 } \right)\).

1. Show that the equation of the normal to \(C\) at \(A\) can be written as \(10 y = 4 x - 27\) The normal to \(C\) at \(A\) meets \(C\) again at the point \(B\), as shown in Figure 3.
2. Use algebra to find the coordinates of \(B\).

1. (i) Find \(\frac { \mathrm { d } } { \mathrm { d } x } \ln \left( \sin ^ { 2 } 3 x \right)\) writing your answer in simplest form.
   (ii) (a) Find \(\frac { \mathrm { d } } { \mathrm { d } x } \left( 3 x ^ { 2 } - 4 \right) ^ { 6 }\) (b) Hence show that

$$\int _ { 0 } ^ { \sqrt { 2 } } x \left( 3 x ^ { 2 } - 4 \right) ^ { 5 } \mathrm {~d} x = R$$ where \(R\) is an integer to be found.
(Solutions relying on calculator technology are not acceptable.)

6. (i) The curve \(C _ { 1 }\) has equation $$y = 3 \ln \left( x ^ { 2 } - 5 \right) - 4 x ^ { 2 } + 15 \quad x > \sqrt { 5 }$$ Show that \(C _ { 1 }\) has a stationary point at \(x = \frac { \sqrt { p } } { 2 }\) where \(p\) is a constant to be found.
(ii) A different curve \(C _ { 2 }\) has equation $$y = 4 x - 12 \sin ^ { 2 } x$$

1. Show that, for this curve, $$\frac { \mathrm { d } y } { \mathrm {~d} x } = A + B \sin 2 x$$ where \(A\) and \(B\) are constants to be found.
2. Hence, state the maximum gradient of this curve.

8. A curve \(C\) has equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \arcsin \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2 \quad - \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }$$

1. Sketch \(C\).
2. Given \(x = 2 \sin y\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { A - x ^ { 2 } } }$$ where \(A\) is a constant to be found. The point \(P\) lies on \(C\) and has \(y\) coordinate \(\frac { \pi } { 4 }\)
3. Find the equation of the tangent to \(C\) at \(P\). Write your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants to be found.
   (3)

4. $$y = \log _ { 10 } ( 2 x + 1 )$$

1. Express \(x\) in terms of \(y\).
2. Hence, giving your answer in terms of \(x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) 4.
   1. Express \(x\) in terms of \(y\).

      	\(\begin{array} { c } \text { Leave }
      \text { blank } \end{array}\)

      (2)
   2. Hence, giving your answer in terms of \(x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) \includegraphics[max width=\textwidth, alt={}, center]{83e12fa4-1abb-4bea-bff4-8d36757bd9c3-10_2662_111_107_1950}

1. The curve \(C\) has equation

$$y = \frac { \ln \left( x ^ { 2 } + k \right) } { x ^ { 2 } + k } \quad x \in \mathbb { R }$$ where \(k\) is a positive constant.

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A x \left( B - \ln \left( x ^ { 2 } + k \right) \right) } { \left( x ^ { 2 } + k \right) ^ { 2 } }$$ where \(A\) and \(B\) are constants to be found. Given that \(C\) has exactly three turning points,
2. find the \(x\) coordinate of each of these points. Give your answer in terms of \(k\) where appropriate.
3. find the upper limit to the value for \(k\).

10. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$ Point \(A\) is the minimum turning point on the curve.

1. Show, by using calculus, that the \(x\) coordinate of point \(A\) is a solution of $$x = \frac { 6 } { 1 + \ln \left( x ^ { 2 } \right) }$$
2. Starting with \(x _ { 0 } = 2.27\), use the iteration $$x _ { n + 1 } = \frac { 6 } { 1 + \ln \left( x _ { n } ^ { 2 } \right) }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. Use your answer to part (b) to deduce the coordinates of point \(A\) to one decimal place.

1. (i) Differentiate \(y = 5 x ^ { 2 } \ln 3 x , \quad x > 0\) (ii) Given that

$$y = \frac { x } { \sin x + \cos x } , \quad - \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 }$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 1 + x ) \sin x + ( 1 - x ) \cos x } { 1 + \sin 2 x } , \quad - \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 }$$ \includegraphics[max width=\textwidth, alt={}, center]{e30f0c28-1695-40a1-8e9a-6ea7e29042bf-11_99_104_2631_1781}

13. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve \(C\) with equation $$y = \frac { 1 } { 2 x } \ln 2 x , \quad x > \frac { 1 } { 2 }$$ The finite region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = \mathrm { e }\) and \(x = 5 \mathrm { e }\). The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { 2 x } \ln 2 x\). The values for \(y\) are given to 4 significant figures.

1. Use the trapezium rule with all the \(y\) values in the table to find an approximate value for the area of \(R\), giving your answer to 3 significant figures.
2. Using the substitution \(u = \ln 2 x\), or otherwise, find \(\int \frac { 1 } { 2 x } \ln 2 x \mathrm {~d} x\)
3. Use your answer to part (b) to find the true area of \(R\), giving your answer to 3 significant figures.
4. Using calculus, find an equation for the tangent to the curve at the point where \(x = \frac { \mathrm { e } ^ { 2 } } { 2 }\), giving your answer in the form \(y = m x + c\) where \(m\) and \(c\) are exact multiples of powers of e.

10. (a) Use the identity for \(\sin ( A + B )\) to prove that $$\sin 2 A \equiv 2 \sin A \cos A$$ (b) Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ \ln \left( \tan \left( \frac { 1 } { 2 } x \right) \right) \right] = \operatorname { cosec } x$$ A curve \(C\) has the equation $$y = \ln \left( \tan \left( \frac { 1 } { 2 } x \right) \right) - 3 \sin x , \quad 0 < x < \pi$$ (c) Find the \(x\) coordinates of the points on \(C\) where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) Give your answers to 3 decimal places.
(Solutions based entirely on graphical or numerical methods are not acceptable.)

5. (i) Find the \(x\) coordinate of each point on the curve \(y = \frac { x } { x + 1 } , x \neq - 1\), at which the gradient is \(\frac { 1 } { 4 }\) (ii) Given that $$\int _ { a } ^ { 2 a } \frac { t + 1 } { t } \mathrm {~d} t = \ln 7 \quad a > 0$$ find the exact value of the constant \(a\).

1. (a) Prove by differentiation that

$$\frac { \mathrm { d } } { \mathrm {~d} y } ( \ln \tan 2 y ) = \frac { 4 } { \sin 4 y } , \quad 0 < y < \frac { \pi } { 4 }$$ (b) Given that \(y = \frac { \pi } { 6 }\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos x \sin 4 y , \quad 0 < y < \frac { \pi } { 4 }$$ Give your answer in the form \(\tan 2 y = A \mathrm { e } ^ { B \sin x }\), where \(A\) and \(B\) are constants to be determined.

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = 2 \mathrm { e } ^ { - x } \sqrt { \sin x } , 0 \leqslant x \leqslant \pi\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve and the \(x\)-axis.

1. Complete the table below with the value of \(y\) corresponding to \(x = \frac { \pi } { 2 }\), giving your answer to 5 decimal places.

   \(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
   \(y\)	0	0.76679		0.15940	0

2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of the region \(R\). Give your answer to 4 decimal places.
3. Given \(y = 2 \mathrm { e } ^ { - x } \sqrt { \sin x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) for \(0 < x < \pi\). The curve \(C\) has a maximum turning point when \(x = a\).
4. Use your answer to part (c) to find the value of \(a\), giving your answer to 3 decimal places.

7. A curve has equation $$y = \ln ( 1 - \cos 2 x ) , \quad x \in \mathbb { R } , 0 < x < \pi$$ Show that

1. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = k \cot x\), where \(k\) is a constant to be found. Hence find the exact coordinates of the point on the curve where
2. \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { 3 }\)

3. The point \(P\) lies on the curve with equation \(y = \ln \left( \frac { 1 } { 3 } x \right)\). The \(x\)-coordinate of \(P\) is 3 . Find an equation of the normal to the curve at the point \(P\) in the form \(y = a x + b\), where \(a\) and \(b\) are constants.
(5)

4. (i) Given that \(y = \frac { \ln \left( x ^ { 2 } + 1 \right) } { x }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
(ii) Given that \(x = \tan y\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }\).

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 1.

1. Write down the coordinates of \(A\) and the coordinates of \(B\).
2. Find f'(x).
3. Show that the \(x\)-coordinate of \(Q\) lies between 3.5 and 3.6
4. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \frac { 8 } { 1 + \ln x }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } }$$ is used.
5. Taking \(x _ { 0 } = 3.55\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

Differentiate with respect to \(x\), giving your answer in its simplest form,

1. \(x ^ { 2 } \ln ( 3 x )\)
2. \(\frac { \sin 4 x } { x ^ { 3 } }\)

2. (a) Differentiate with respect to \(x\)

1. \(3 \sin ^ { 2 } x + \sec 2 x\),
2. \(\{ x + \ln ( 2 x ) \} ^ { 3 }\). Given that \(y = \frac { 5 x ^ { 2 } - 10 x + 9 } { ( x - 1 ) ^ { 2 } } , \quad x \neq 1\),
   (b) show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 8 } { ( x - 1 ) ^ { 3 } }\).

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.