1.07q Product and quotient rules: differentiation

9 The equation of a curve is \(y = 3 x + 1 - 4 ( 3 x + 1 ) ^ { \frac { 1 } { 2 } }\) for \(x > - \frac { 1 } { 3 }\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
2. Find the coordinates of the stationary point of the curve and determine its nature.

5 A curve has the equation \(\mathrm { y } = \frac { 3 } { 2 \mathrm { x } ^ { 2 } - 5 }\).
Find the equation of the normal to the curve at the point \(( 2,1 )\), giving your answer in the form \(\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0\), where \(a , b\) and \(c\) are integers.

6 The equation of a curve is \(y = 2 + \sqrt { 25 - x ^ { 2 } }\).
Find the coordinates of the point on the curve at which the gradient is \(\frac { 4 } { 3 }\).

1 It is given that \(\mathrm { f } ( x ) = ( 2 x - 5 ) ^ { 3 } + x\), for \(x \in \mathbb { R }\). Show that f is an increasing function.

8 The equation of a curve is \(y = ( 2 x - 3 ) ^ { 3 } - 6 x\).

1. Express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) in terms of \(x\).
2. Find the \(x\)-coordinates of the two stationary points and determine the nature of each stationary point.

5 A curve has equation \(y = 2 x + \frac { 1 } { ( x - 1 ) ^ { 2 } }\). Verify that the curve has a stationary point at \(x = 2\) and determine its nature.

5 \includegraphics[max width=\textwidth, alt={}, center]{8bdd1285-9e39-465a-8c09-bbe410504f9d-06_442_698_260_721} The diagram shows part of the curve with equation \(y = x ^ { 3 } \cos 2 x\). The curve has a maximum at the point \(M\).

1. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \sqrt [ 3 ] { 1.5 x ^ { 2 } \cot 2 x }\).
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(M\) lies between 0.59 and 0.60.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(M\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

2 Find the exact coordinates of the stationary point on the curve with equation \(y = 5 x \mathrm { e } ^ { \frac { 1 } { 2 } x }\).

5 \includegraphics[max width=\textwidth, alt={}, center]{2d6fc4c5-70ec-4cd8-9b48-59d5ce0e39b7-08_575_618_262_762} The diagram shows the curve with equation \(y = \frac { 3 x + 2 } { \ln x }\). The curve has a minimum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { 3 x + 2 } { 3 \ln x }\). [3]
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(M\) lies between 3 and 4.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(M\) correct to 5 significant figures. Give the result of each iteration to 7 significant figures.

6 A curve has equation \(y = \frac { 9 \mathrm { e } ^ { 2 x } + 16 } { \mathrm { e } ^ { x } - 1 }\).

1. Show that the \(x\)-coordinate of any stationary point on the curve satisfies the equation $$\mathrm { e } ^ { x } \left( 3 \mathrm { e } ^ { x } - 8 \right) \left( 3 \mathrm { e } ^ { x } + 2 \right) = 0$$
2. Hence show that the curve has only one stationary point and find its exact coordinates.

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

5 \includegraphics[max width=\textwidth, alt={}, center]{a1ea242a-c7f4-46b0-b4b8-bd13b3880557-06_526_947_276_591} The diagram shows the curve with equation \(y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } \left( x ^ { 2 } - 5 x + 4 \right)\). The curve crosses the \(x\)-axis at the points \(A\) and \(B\), and has a maximum at the point \(C\).

1. Find the exact gradient of the curve at \(B\).
2. Find the exact coordinates of \(C\).

5 A curve has equation \(\mathrm { y } = \frac { 1 + \mathrm { e } ^ { 2 \mathrm { x } } } { 1 + 3 \mathrm { x } }\). The curve has exactly one stationary point \(P\).

1. Find \(\frac { \mathrm { dy } } { \mathrm { dx } }\) and hence show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 6 } + \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }\).
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.35 and 0.45 .
3. Use an iterative formula based on the equation in part (a) to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{971a1d8d-a82e-4a3a-b72d-3c147e4f30bb-10_451_647_258_699} The diagram shows the curve with equation \(\mathrm { y } = \sqrt { \sin 2 \mathrm { x } + \sin ^ { 2 } 2 \mathrm { x } }\) for \(0 \leqslant x \leqslant \frac { 1 } { 6 } \pi\). The shaded region is bounded by the curve and the straight lines \(x = \frac { 1 } { 6 } \pi\) and \(y = 0\).

6 \includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-10_417_700_310_685} The diagram shows the curve with equation \(y = \frac { \ln ( 2 x + 1 ) } { x + 3 }\). The curve has a maximum point \(M\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \frac { x + 3 } { \ln ( 2 x + 1 ) } - 0.5\).
3. Show by calculation that the \(x\)-coordinate of \(M\) lies between 2.5 and 3.0 .
4. Use an iterative formula based on the equation in part (b) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

6 A curve has equation \(y = x ^ { 3 } \mathrm { e } ^ { 0.2 x }\) where \(x \geqslant 0\). At the point \(P\) on the curve, the gradient of the curve is 15 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt { \frac { 75 \mathrm { e } ^ { - 0.2 x } } { 15 + x } }\).
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(P\) lies between 1.7 and 1.8.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 \includegraphics[max width=\textwidth, alt={}, center]{9cf008d5-c15f-4491-9e4d-4bd070f896d5-06_446_832_260_653} The diagram shows part of the curve with equation \(y = \frac { 5 x } { 4 x ^ { 3 } + 1 }\). The shaded region is bounded by the curve and the lines \(x = 1 , x = 3\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of the maximum point.
2. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures.
3. State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region.

8 A curve has equation \(y = f ( x )\) where \(f ( x ) = \frac { 4 x ^ { 3 } + 8 x - 4 } { 2 x - 1 }\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the coordinates of each of the stationary points of the curve \(y = \mathrm { f } ( x )\).
2. Divide \(4 x ^ { 3 } + 8 x - 4\) by ( \(2 x - 1\) ), and hence find \(\int \mathrm { f } ( x ) \mathrm { d } x\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

4 \includegraphics[max width=\textwidth, alt={}, center]{c473f577-1e96-4d11-a0d5-cdfa4873c295-06_460_1445_260_349} The diagram shows the curve with equation \(y = \frac { x - 2 } { x ^ { 2 } + 8 }\). The shaded region is bounded by the curve and the lines \(x = 14\) and \(y = 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence determine the exact \(x\)-coordinates of the stationary points.
2. Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures.

7 \includegraphics[max width=\textwidth, alt={}, center]{1cd04df5-3fe3-4573-b880-d49262afd16a-10_456_598_269_762} The diagram shows the curve with equation \(y = \frac { 2 \ln x } { 3 x + 1 }\). The curve crosses the \(x\)-axis at the point \(A\) and has a maximum point \(B\). The shaded region is bounded by the curve and the lines \(x = 3\) and \(y = 0\).

1. Find the gradient of the curve at \(A\).
2. Show by calculation that the \(x\)-coordinate of \(B\) lies between 3.0 and 3.1.
3. Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places.

2 A curve has equation \(y = 3 \tan \frac { 1 } { 2 } x \cos 2 x\).
Find the gradient of the curve at the point for which \(x = \frac { 1 } { 3 } \pi\).

5 The equation of a curve is \(y = x ^ { 3 } \mathrm { e } ^ { - x }\).

1. Show that the curve has a stationary point where \(x = 3\).
2. Find the equation of the tangent to the curve at the point where \(x = 1\).

5 Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = 4\) in each of the following cases:

1. \(y = x \ln ( x - 3 )\),
2. \(y = \frac { x - 1 } { x + 1 }\).

6 The equation of a curve is \(y = \frac { 3 x ^ { 2 } } { x ^ { 2 } + 4 }\). At the point on the curve with positive \(x\)-coordinate \(p\), the gradient of the curve is \(\frac { 1 } { 2 }\).

1. Show that \(p = \sqrt { } \left( \frac { 48 p - 16 } { p ^ { 2 } + 8 } \right)\).
2. Show by calculation that \(2 < p < 3\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

8 \includegraphics[max width=\textwidth, alt={}, center]{6295873e-7db4-4e7e-8dcd-912ad9c41675-10_643_414_260_863} The diagram shows the curve with equation $$y = 3 x ^ { 2 } \ln \left( \frac { 1 } { 6 } x \right) .$$ The curve crosses the \(x\)-axis at the point \(P\) and has a minimum point \(M\).

1. Find the gradient of the curve at the point \(P\).
2. Find the exact coordinates of the point \(M\).

4 Find the equation of the tangent to the curve \(y = \frac { \mathrm { e } ^ { 4 x } } { 2 x + 3 }\) at the point on the curve for which \(x = 0\). Give your answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.