1.07q Product and quotient rules: differentiation

2. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { x } \left( 2 y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } + 1 \right)$$

1. Show that $$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \mathrm { e } ^ { x } \left[ 2 y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + k y \frac { \mathrm {~d} y } { \mathrm {~d} x } + y ^ { 2 } + 1 \right]$$ where \(k\) is a constant to be found. Given that, at \(x = 0 , y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\),
2. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).

5. $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 x + y ^ { 2 }$$

1. Show that $$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 1 - 2 y ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 3$$ Given that \(y = 1\) at \(x = 1\),
2. find a series solution for \(y\) in ascending powers of ( \(x - 1\) ), up to and including the term in \(( x - 1 ) ^ { 3 }\).

1. Given that

$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } + 5 y = 0$$

1. find \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) in terms of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } , \frac { \mathrm {~d} y } { \mathrm {~d} x }\) and \(y\). Given that \(y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\) at \(x = 0\)
2. find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).

1. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 2 \cos x$$

1. Find \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) in terms of \(x , \frac { \mathrm {~d} y } { \mathrm {~d} x }\) and \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\). At \(x = 0 , y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\)
2. Find the value of \(\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }\) at \(x = 0\)
3. Express \(y\) as a series in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).

1. Given that

$$y = 3 x \arcsin 2 x \quad 0 \leqslant x \leqslant \frac { 1 } { 2 }$$

1. determine an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
2. Hence determine the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = \frac { 1 } { 4 }\), giving your answer in the form \(a \pi + b\) where \(a\) and \(b\) are fully simplified constants to be found.

7 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in each of the following cases:

1. \(y = \frac { 1 } { 2 } x ^ { 4 } - 3 x\),
2. \(y = \left( 2 x ^ { 2 } + 3 \right) ( x + 1 )\),
3. \(y = \sqrt [ 5 ] { x }\).

7 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 \(\int _ { 0 } ^ { 1 } 2 x \ln ( 1 + x ) \mathrm { d } x\).

8 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 \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

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. Find the coordinates of the stationary points of the curve with equation $$y = \frac { x - 1 } { x ^ { 2 } - 2 x + 5 }$$

1. $$f ( x ) = \frac { 4 x - 1 } { 2 x + 1 }$$ Find an equation for the tangent to the curve \(y = \mathrm { f } ( x )\) at the point where \(x = - 2\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

7. $$\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { 4 x + 1 } , \quad x \in \mathbb { R } , \quad x \neq - \frac { 1 } { 4 }$$

1. Find and simplify an expression for \(\mathrm { f } ^ { \prime } ( x )\).
2. Find the set of values of \(x\) for which \(\mathrm { f } ( x )\) is increasing.
3. Use Simpson's rule with six strips to find an approximate value for $$\int _ { 0 } ^ { 6 } f ( x ) d x$$

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.

3

1. Differentiate \(x ^ { 2 } ( x + 1 ) ^ { 6 }\) with respect to \(x\).
2. Find the gradient of the curve \(y = \frac { x ^ { 2 } + 3 } { x ^ { 2 } - 3 }\) at the point where \(x = 1\).

8 \includegraphics[max width=\textwidth, alt={}, center]{1216a06e-7e14-48d7-a7ca-7acd8d71af5f-4_538_1443_262_351} The diagram shows the curve with equation \(y = x ^ { 8 } \mathrm { e } ^ { - x ^ { 2 } }\). The curve has maximum points at \(P\) and \(Q\). The shaded region \(A\) is bounded by the curve, the line \(y = 0\) and the line through \(Q\) parallel to the \(y\)-axis. The shaded region \(B\) is bounded by the curve and the line \(P Q\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2 .
2. Use Simpson's rule with 4 strips to find an approximation to the area of region \(A\). Give your answer correct to 3 decimal places.
3. Deduce an approximation to the area of region \(B\).

7 A curve has equation \(y = \frac { x \mathrm { e } ^ { 2 x } } { x + k }\), where \(k\) is a non-zero constant.

1. Differentiate \(x \mathrm { e } ^ { 2 x }\), and show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \mathrm { e } ^ { 2 x } \left( 2 x ^ { 2 } + 2 k x + k \right) } { ( x + k ) ^ { 2 } }\).
2. Given that the curve has exactly one stationary point, find the value of \(k\), and determine the exact coordinates of the stationary point.

6

1. Find the exact value of the \(x\)-coordinate of the stationary point of the curve \(y = x \ln x\).
2. The equation of a curve is \(y = \frac { 4 x + c } { 4 x - c }\), where \(c\) is a non-zero constant. Show by differentiation that this curve has no stationary points.

1 Differentiate each of the following with respect to \(x\).

1. \(x ^ { 3 } ( x + 1 ) ^ { 5 }\)
2. \(\sqrt { 3 x ^ { 4 } + 1 }\)

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.

5 The equation of a curve is \(y = \frac { x ^ { 2 } } { 2 x + 1 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( x + 1 ) } { ( 2 x + 1 ) ^ { 2 } }\).
2. Find the coordinates of the stationary points of the curve. You need not determine their nature.

8 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \ln ( 1 + \sqrt { x } )\) for \(x \geqslant 0\).
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).

8 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.

3

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

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.