Find gradient at a point - direct evaluation

3 The equation of a curve is \(y = \frac { 2 } { \sqrt { } ( 5 x - 6 ) }\).

1. Find the gradient of the curve at the point where \(x = 2\).
2. Find \(\int \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x\) and hence evaluate \(\int _ { 2 } ^ { 3 } \frac { 2 } { \sqrt { } ( 5 x - 6 ) } \mathrm { d } x\).

3
\includegraphics[max width=\textwidth, alt={}, center]{b104e2a7-06c8-4e2e-a4f9-5095ad56897a-04_652_392_274_872} The diagram shows the curve with equation \(y = 6 \mathrm { e } ^ { - \frac { 1 } { 2 } x }\). The points on the curve with \(x\)-coordinates 0 and 2 are denoted by \(A\) and \(B\) respectively. The shaded region is enclosed by the curve, the line through \(A\) parallel to the \(x\)-axis and the line through \(B\) parallel to the \(y\)-axis.

1. Find the exact gradient of the curve at \(B\).
2. Find the exact area of the shaded region.

3 The function f is defined by \(\mathrm { f } ( x ) = \tan ^ { 2 } \left( \frac { 1 } { 2 } x \right)\) for \(0 \leqslant x < \pi\).

1. Find the exact value of \(\mathrm { f } ^ { \prime } \left( \frac { 2 } { 3 } \pi \right)\).
   \includegraphics[max width=\textwidth, alt={}, center]{dcc483e9-630e-4f02-ad8c-4a27c0720fc6-05_2726_33_97_22}
2. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } ( \mathrm { f } ( x ) + \sin x ) \mathrm { d } x\).

2 Let \(\mathrm { f } ( x ) = 4 \sin ^ { 2 } 3 x\).

1. Find the value of \(\mathrm { f } ^ { \prime } \left( \frac { 1 } { 4 } \pi \right)\).
2. Find \(\int \mathrm { f } ( x ) \mathrm { d } x\).
   \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-05_2723_35_101_20}

3 The function f is defined by \(\mathrm { f } ( x ) = \tan ^ { 2 } \left( \frac { 1 } { 2 } x \right)\) for \(0 \leqslant x < \pi\).

1. Find the exact value of \(\mathrm { f } ^ { \prime } \left( \frac { 2 } { 3 } \pi \right)\).
   \includegraphics[max width=\textwidth, alt={}, center]{468efb3f-be7b-4f9e-b8c3-c6fd40d7714a-05_2726_33_97_22}
2. Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } ( \mathrm { f } ( x ) + \sin x ) \mathrm { d } x\).

2 Find the gradient of each of the following curves at the point for which \(x = 0\).

1. \(y = 3 \sin x + \tan 2 x\)
2. \(y = \frac { 6 } { 1 + \mathrm { e } ^ { 2 x } }\)

1 Find the gradient of the curve $$y = 3 e ^ { 4 x } - 6 \ln ( 2 x + 3 )$$ at the point for which \(x = 0\).

5 The expression \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 3 x \mathrm { e } ^ { - 2 x }\).

1. Find the exact value of \(\mathrm { f } ^ { \prime } \left( - \frac { 1 } { 2 } \right)\).
2. Find the exact value of \(\int _ { - \frac { 1 } { 2 } } ^ { 0 } \mathrm { f } ( x ) \mathrm { d } x\).

1 Find the gradient of the curve \(y = \ln ( 5 x + 1 )\) at the point where \(x = 4\).

7. (i) Given that \(y = \tan x + 2 \cos x\), find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = \frac { \pi } { 4 }\).
(ii) Given that \(x = \tan \frac { 1 } { 2 } y\), prove that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { 1 + x ^ { 2 } }\).
(iii) Given that \(y = \mathrm { e } ^ { - x } \sin 2 x\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) can be expressed in the form \(R \mathrm { e } ^ { - x } \cos ( 2 x + \alpha )\). Find, to 3 significant figures, the values of \(R\) and \(\alpha\), where \(0 < \alpha < \frac { \pi } { 2 }\).

9 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 \(y = \mathrm { f } ( x )\) ?

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

1 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 $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
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 \(\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x\), 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 \(x = 1\).

2 Find the exact gradient of the curve \(y = \ln ( 1 - \cos 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\).

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 The function \(f ( x ) = \ln \left( t + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\).
4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
   Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

3 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 \(y = \mathrm { f } ( x )\) ?

1 At the point ( 1,0 ) on the curve \(y = \ln x\), which statement below is correct? Tick ( \(\checkmark\) ) one box. The gradient is negative and decreasing □ The gradient is negative and increasing
\includegraphics[max width=\textwidth, alt={}, center]{091aecd0-d812-4a8f-8596-a1c91f3bae1c-02_109_109_995_1306} The gradient is positive and decreasing □ The gradient is positive and increasing □

1 Find the gradient of the curve \(y = \mathrm { e } ^ { - 3 x }\) at the point where it crosses the \(y\)-axis. Circle your answer.
\(\begin{array} { l l l } - 3 & - 1 & 1 \end{array}\)

3 Find the gradient of the tangent to the curve $$y = \sin ^ { - 1 } x$$ at the point where \(x = \frac { 1 } { 5 }\)
Circle your answer.