1.07q Product and quotient rules: differentiation

3 Differentiate the following functions.

1. \(\quad y = \left( x ^ { 2 } + 3 \right) ^ { 5 }\)
2. \(y = \frac { \sin 2 x } { x }\)

8 A curve has equation \(y = ( x + 2 ) \mathrm { e } ^ { - x }\).

1. Find the coordinates of the points where the curve cuts the axes.
2. Find the coordinates of the stationary point, S , on the curve.
3. By evaluating \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at S , determine whether the stationary point is a maximum or a minimum.
4. Sketch the curve in the domain \(- 3 < x < 3\).
5. Find where the normal to the curve at the point \(( 0,2 )\) cuts the curve again.
6. Find the area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

8 Fig. 8 shows part of the graph of the function \(y = 5 x ( 2 x - 1 ) ^ { 3 }\). \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of S , the turning point of the curve.
2. Find the area of the shaded region enclosed between the curve and the \(x\)-axis.
3. Given that \(\mathrm { f } ( x ) = 5 x ( 2 x - 1 ) ^ { 3 }\), show that \(\mathrm { f } ( x + 0.5 ) = 40 x ^ { 3 } ( x + 0.5 )\).
4. Find \(\int _ { - \frac { 1 } { 2 } } ^ { 0 } 40 x ^ { 3 } ( x + 0.5 ) \mathrm { d } x\).
5. Explain, with the aid of a sketch, the connection between your answer to parts (ii) and (iv).

4 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
4. Find the exact area of the shaded region between the line OP and the curve.

1 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h] \end{figure}

1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
2. Find the exact coordinates of the turning point P .
3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\) - and \(y\)-axes.

1

1. Differentiate \(\sqrt { 1 + 3 x ^ { 2 } }\).
2. Hence show that the derivative of \(x \sqrt { 1 + 3 x ^ { 2 } }\) is \(\frac { 1 + 6 x ^ { 2 } } { \sqrt { 1 + 3 x ^ { 2 } } }\).

4

1. Differentiate \(\frac { \ln x } { x ^ { 2 } }\), simplifying your answer.
2. Using integration by parts, show that \(\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c\).

1 Fig. 1 shows part of the curve \(y = \mathrm { e } ^ { 2 x } \cos x\). \begin{figure}[h] \end{figure} Find the coordinates of the turning point P .

3

1. Given that \(y = \mathrm { e } ^ { - x } \sin 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence show that the curve \(y = \mathrm { e } ^ { - x } \sin 2 x\) has a stationary point when \(x = \frac { 1 } { 2 } \arctan 2\).

5 Differentiate \(x ^ { 2 } \tan 2 x\).

7 Given that \(y = x ^ { 2 } \sqrt { 1 + 4 x }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( 5 x + 1 ) } { \sqrt { 1 + 4 x } }\).

3

1. Differentiate \(x \cos 2 x\) with respect to \(x\).
2. Integrate \(x \cos 2 x\) with respect to \(x\).

1 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

2 Fig. 9 shows the curve \(y = x \mathrm { e } ^ { - 2 x }\) together with the straight line \(y = m x\), where \(m\) is a constant, with \(0 < m < 1\). The curve and the line meet at O and P . The dashed line is the tangent at P . \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of P is \(- \frac { 1 } { 2 } \ln m\).
2. Find, in terms of \(m\), the gradient of the tangent to the curve at P . You are given that OP and this tangent are equally inclined to the \(x\)-axis.
3. Show that \(m = \mathrm { e } ^ { - 2 }\), and find the exact coordinates of P .
4. Find the exact area of the shaded region between the line OP and the curve. END OF QUESTION PAPER

3

1. Use the substitution \(u = 1 + x\) to show that $$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$ where \(a\) and \(b\) are to be found.
   Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x\), giving your answer in exact form. Fig. 8 shows the curve \(y = x ^ { 2 } \ln ( 1 + x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure}
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
3. Using integration by parts, and the result of part (i), find the exact area enclosed by the curve \(y = x ^ { 2 } \ln ( 1 + x )\), the \(x\)-axis and the line \(x = 1\).

1 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.

2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P . \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

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

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence verify that the origin is a stationary point of the curve.
2. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), and use this to verify that the origin is a minimum point.
3. Using the substitution \(u = 1 + x\), show that \(\int \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x = \int \left( u - 2 + \frac { 1 } { u } \right) \mathrm { d } u\). Hence evaluate \(\int _ { 0 } ^ { 1 } \frac { x ^ { 2 } } { 1 + x } \mathrm {~d} x\), giving your answer in an exact form.
4. Using integration by parts and your answer to part (iii), evaluate \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

2 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

3 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).

1 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h] \end{figure}

1. Find the exact coordinates of P .
2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.

2 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \cos ^ { 2 } x } , - \frac { 1 } { 2 } \pi < x < \frac { 1 } { 2 } \pi\), together with its asymptotes \(x = \frac { 1 } { 2 } \pi\) and \(x = - \frac { 1 } { 2 } \pi\). \begin{figure}[h] \end{figure}

1. Use the quotient rule to show that the derivative of \(\frac { \sin x } { \cos x }\) is \(\frac { 1 } { \cos ^ { 2 } x }\).
2. Find the area bounded by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 4 } \pi\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \mathrm { f } \left( x + \frac { 1 } { 4 } \pi \right)\).
3. Verify that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) cross at \(( 0,1 )\).
4. State a sequence of two transformations such that the curve \(y = \mathrm { f } ( x )\) is mapped to the curve \(y = \mathrm { g } ( x )\). On the copy of Fig. 9, sketch the curve \(y = \mathrm { g } ( x )\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve.
5. Use your result from part (ii) to write down the area bounded by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = - \frac { 1 } { 4 } \pi\).

1 Fig. 8 shows a sketch of part of the curve \(y = x \sin 2 x\), where \(x\) is in radians.
The curve crosses the \(x\)-axis at the point P . The tangent to the curve at P crosses the \(y\)-axis at Q . \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that the \(x\)-coordinates of the turning points of the curve satisfy the equation \(\tan 2 x + 2 x = 0\).
2. Find, in terms of \(\pi\), the \(x\)-coordinate of the point P . Show that the tangent PQ has equation \(2 \pi x + 2 y = \pi ^ { 2 }\).
   Find the exact coordinates of Q.
3. Show that the exact value of the area shaded in Fig. 8 is \(\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)\).