1.07q Product and quotient rules: differentiation

Differentiate with respect to \(x\)

1. \(x^4 + 6\sqrt{x}\), [3]
2. \(\frac{(x + 4)^3}{x}\). [4]

1. Use integration by parts to show that $$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = \frac{1}{4}\pi - \frac{1}{2}\ln 2.$$ [6]

\includegraphics{figure_1} The finite region \(R\), bounded by the equation \(y = x^{\frac{1}{2}} \sec x\), the line \(x = \frac{\pi}{4}\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Find the volume of the solid of revolution generated. [2]
2. Find the gradient of the curve with equation \(y = x^{\frac{1}{2}} \sec x\) at the point where \(x = \frac{\pi}{4}\). [3]

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

1. Find the values of the constants \(A\), \(B\) and \(C\) such that $$\text{f}(x) = Ax^{\frac{3}{2}} + Bx^{\frac{1}{2}} + Cx^{-\frac{1}{2}}.$$ [3]
2. Show that $$\text{f}'(x) = \frac{(3x+4)(x-4)}{4x^{\frac{3}{2}}}.$$ [6]

A curve has the equation \(y = (\sqrt{x} - 3)^2\), \(x \geq 0\).

1. Show that \(\frac{dy}{dx} = 1 - \frac{3}{\sqrt{x}}\). [4]

The point \(P\) on the curve has \(x\)-coordinate 4.

1. Find an equation for the normal to the curve at \(P\) in the form \(y = mx + c\). [5]
2. Show that the normal to the curve at \(P\) does not intersect the curve again. [4]

Differentiate with respect to \(x\) $$\frac{6x^2 - 1}{2\sqrt{x}}.$$ [5]

Use the derivatives of \(\sin x\) and \(\cos x\) to prove that the derivative of \(\tan x\) is \(\sec^2 x\). [4]

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

\includegraphics{figure_8} The diagram shows the curve with equation \(y = x^8 e^{-x^2}\). 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, the line \(y = 0\) and the line \(PQ\).

1. Show by differentiation that the \(x\)-coordinate of \(Q\) is 2. [5]
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. [4]
3. Deduce an approximation to the area of region \(B\). [2]

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

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

For each of the following curves, find the gradient at the point with \(x\)-coordinate 2.

1. \(y = \frac{3x}{2x + 1}\) [3]
2. \(y = \sqrt{4x^2 + 9}\) [3]

Find \(\frac{dy}{dx}\) in each of the following cases:

1. \(y = x^3 e^{2x}\), [2]
2. \(y = \ln(3 + 2x^2)\), [2]
3. \(y = \frac{x}{2x + 1}\). [2]

Fig. 8 shows part of the curve \(y = x \sin 3x\). It crosses the \(x\)-axis at P. The point on the curve with \(x\)-coordinate \(\frac{1}{6}\pi\) is Q. \includegraphics{figure_8}

1. Find the \(x\)-coordinate of P. [3]
2. Show that Q lies on the line \(y = x\). [1]
3. Differentiate \(x \sin 3x\). Hence prove that the line \(y = x\) touches the curve at Q. [6]
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac{1}{72}(\pi^2 - 8)\). [7]

Fig. 9 shows the curve \(y = f(x)\), where \(f(x) = \frac{1}{\cos^2 x}\), \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), together with its asymptotes \(x = \frac{1}{2}\pi\) and \(x = -\frac{1}{2}\pi\). \includegraphics{figure_9}

1. Use the quotient rule to show that the derivative of \(\frac{\sin x}{\cos x}\) is \(\frac{1}{\cos^2 x}\). [3]
2. Find the area bounded by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac{1}{4}\pi\). [3]

The function \(g(x)\) is defined by \(g(x) = \frac{1}{2}f(x + \frac{1}{4}\pi)\).

1. Verify that the curves \(y = f(x)\) and \(y = g(x)\) cross at \((0, 1)\). [3]
2. State a sequence of two transformations such that the curve \(y = f(x)\) is mapped to the curve \(y = g(x)\). On the copy of Fig. 9, sketch the curve \(y = g(x)\), indicating clearly the coordinates of the minimum point and the equations of the asymptotes to the curve. [8]
3. Use your result from part (ii) to write down the area bounded by the curve \(y = g(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = -\frac{1}{4}\pi\). [1]

Differentiate \(x^2 \tan 2x\). [3]

Fig. 8 shows parts of the curves \(y = f(x)\) and \(y = g(x)\), where \(f(x) = \tan x\) and \(g(x) = 1 + f(x - \frac{1}{4}\pi)\). \includegraphics{figure_8}

1. Describe a sequence of two transformations which maps the curve \(y = f(x)\) to the curve \(y = g(x)\). [4]

It can be shown that \(g(x) = \frac{2\sin x}{\sin x + \cos x}\).

1. Show that \(g'(x) = \frac{2}{(\sin x + \cos x)^2}\). Hence verify that the gradient of \(y = g(x)\) at the point \((\frac{1}{4}\pi, 1)\) is the same as that of \(y = f(x)\) at the origin. [7]
2. By writing \(\tan x = \frac{\sin x}{\cos x}\) and using the substitution \(u = \cos x\), show that \(\int_0^{\frac{1}{4}\pi} f(x)dx = \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{u}du\). Evaluate this integral exactly. [4]
3. Hence find the exact area of the region enclosed by the curve \(y = g(x)\), the \(x\)-axis and the lines \(x = \frac{1}{4}\pi\) and \(x = \frac{1}{2}\pi\). [2]

Given that \(y = x^2\sqrt{1 + 4x}\), show that \(\frac{dy}{dx} = \frac{2x(5x + 1)}{\sqrt{1 + 4x}}\). [5]

Fig. 8 shows the curve \(y = f(x)\), where \(f(x) = \frac{x}{\sqrt{2 + x^2}}\). \includegraphics{figure_8}

1. Show algebraically that \(f(x)\) is an odd function. Interpret this result geometrically. [3]
2. Show that \(f'(x) = \frac{2}{(2 + x^2)^{\frac{3}{2}}}\). Hence find the exact gradient of the curve at the origin. [5]
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\). [4]
4. 1. Show that if \(y = \frac{x}{\sqrt{2 + x^2}}\), then \(\frac{1}{y^2} = \frac{2}{x^2} + 1\). [2]
   2. Differentiate \(\frac{1}{y^2} = \frac{2}{x^2} + 1\) implicitly to show that \(\frac{dy}{dx} = \frac{2y^3}{x^3}\). Explain why this expression cannot be used to find the gradient of the curve at the origin. [4]