Higher Maths
Differentiation

Before differentiating, we need to express the surd term and the reciprocal term as powers of \(x\):

$$ \begin{eqnarray} y &=& 5x^3 - 4\sqrt{x} + \small \frac{1}{3x}\normalsize \\[6pt] &=& 5x^3 - 4x^{\frac{1}{2}} + \small \frac{1}{3}\normalsize x^{-1} \end{eqnarray} $$

Now that \(y\) is in differentiable form, we are able to differentiate:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& 15x^2 - 2x^{-\frac{1}{2}} - \small \frac{1}{3}\normalsize x^{-2} \\[6pt] &=& 15x^2 - \small\frac{2}{\sqrt{x}}\normalsize - \small \frac{1}{3x^2}\normalsize \\[6pt] \end{eqnarray} $$

Video solution by Clelland Maths 

First we prepare \(f(t)\) for differentiation:

$$ \begin{eqnarray} f(t) &=& t+\small\frac{3}{t}\normalsize \\[6pt] &=& t+3t^{-1} \end{eqnarray} $$

Now we differentiate:

$$ \begin{eqnarray} f'(t) &=& 1-3t^{-2} \\[6pt] &=& 1-\small\frac{3}{t^2}\normalsize \end{eqnarray} $$

Finally, we substitute \(t=2\):

$$ \begin{eqnarray} f'(2) &=& 1-\small\frac{3}{2^2}\normalsize \\[6pt] &=& 1-\small\frac{3}{4}\normalsize \\[6pt] &=& \small\frac{1}{4}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

First we differentiate:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& \small\frac{4}{8}\normalsize x^3 \\[6pt] &=& \small\frac{1}{2}\normalsize x^3 \end{eqnarray} $$

Now we substitute \(x=-2\) to find the gradient of the tangent:

$$ \begin{eqnarray} m_{\textsf{tan}} &=& \small\frac{1}{2}\normalsize(-2)^3 \\[6pt] &=& \small\frac{1}{2}\normalsize(-8) \\[6pt] &=& -\!4 \end{eqnarray} $$

Before we can find the equation of the tangent, we need to know the \(y\)-coordinate of the point of contact:

$$ \begin{eqnarray} y &=& \small\frac{1}{8}\normalsize x^4-5 \\[6pt] &=& \small\frac{1}{8}\normalsize (-2)^4-5 \\[6pt] &=& \small\frac{1}{8}\normalsize (16)-5 \\[6pt] &=& 2-5 \\[6pt] &=& -\!3 \end{eqnarray} $$

Finally, we use the Nat 5 method to find the equation of the straight line through \((-2,\,-3)\) with gradient \(-4\):

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y+3 &=& -\!4(x+2)\\[6pt] y+3 &=& -\!4x-8\\[6pt] y &=& -\!4x-11 \end{eqnarray} $$

Video solution by Clelland Maths 

First we differentiate:

$$ \small\frac{dy}{dx}\normalsize = 3x^2-6x $$

Then we look for \(x\)-values where the derivative equals 9:

$$ \begin{eqnarray} 3x^2-6x &=& 9 \\[6pt] x^2-2x &=& 3 \\[6pt] x^2-2x-3 &=& 0 \\[6pt] (x+1)(x-3) &=& 0 \\[6pt] x+1=0 &\:\ \small\textsf{or}\normalsize\ & x-3=0 \\[6pt] x=-1 &\:\ \small\textsf{or}\normalsize\ & x=3 \\[6pt] \end{eqnarray} $$

This tells us that there are two such points on the curve.

Finally, we find the corresponding \(y\)-values.

When \(x=-1\): $$ \begin{eqnarray} y &=& (-1)^3-3(-1)^2 \\[6pt] &=& -\!1-3 \\[6pt] &=& -\!4 \end{eqnarray} $$

So one point is \((-1,\,-4)\).

When \(x=3\): $$ \begin{eqnarray} y &=& 3^3-3(3^2) \\[6pt] &=& 27-27 \\[6pt] &=& 0 \end{eqnarray} $$

So the other point is \((3,\,0)\).

Video solution by Clelland Maths 

This example needs a firm understanding of trig quadrants from Nat 5 and radian measure from Higher.

First we differentiate, using the standard derivative:

$$ \begin{eqnarray} f'(x) &=& 5(-2\tiny\ \normalsize \text{sin}\tiny\ \normalsize 2x) \\[6pt] &=& -\!10\tiny\ \normalsize \text{sin}\tiny\ \normalsize 2x \end{eqnarray} $$

Then we substitute \(x=\large\frac{5\pi}{6}\normalsize\):

$$ \begin{eqnarray} f'\small\left(\frac{5\pi}{6}\right)\normalsize &=& -\!10\tiny\ \normalsize \text{sin}\tiny\ \normalsize 2\small\left(\frac{5\pi}{6}\right)\normalsize \\[6pt] &=& -\!10\tiny\ \normalsize \text{sin}\tiny\ \normalsize \small\frac{5\pi}{3}\normalsize \end{eqnarray} $$

To complete this, note that \(\large\frac{5\pi}{3}\normalsize\) is a fourth quadrant angle. It's \(2\pi - \large\frac{\pi}{3}\normalsize\small,\) and sine is negative in the "C" quadrant.

$$ \begin{eqnarray} f'\small\left(\frac{5\pi}{6}\right)\normalsize &=& -\!10\tiny\ \normalsize \text{sin}\tiny\ \normalsize \small\frac{5\pi}{3}\normalsize \\[9pt] &=& -\!10\left(-\text{sin}\,\small\frac{\pi}{3}\normalsize\right)\\[9pt] &=& 10\,\text{sin}\,\small\frac{\pi}{3} \\[9pt] &=& 10\left(\small\frac{\sqrt{3}}{2}\normalsize\right)\\[9pt] &=& 5\sqrt{3} \\[9pt] \end{eqnarray} $$

Note: If you find it hard to think naturally in radians, you may substitute \(\large\frac{5\pi}{6}\normalsize=(180\!\div6\!\times 5)^\circ=150^\circ\) instead.

Video solution by Clelland Maths 

First we prepare \(h(x)\) for differentiation:

$$ \begin{eqnarray} h(x) &=& \frac{2}{(1-4x)^5} \\[9pt] &=& 2(1-4x)^{-5} \end{eqnarray} $$

Then we differentiate using the chain rule:

$$ \begin{eqnarray} h'(x) &=& -\!10{(1-4x)^{-6}}(-4) \\[6pt] &=& 40(1-4x)^{-6} \end{eqnarray} $$

Finally, we substitute \(x=\large\frac{1}{8}\normalsize\):

$$ \begin{eqnarray} h'\small\left(\frac{1}{8}\right)\normalsize &=& 40\left(1-4 \small \left( \frac{1}{8} \right) \normalsize\right)^{-6} \\[6pt] &=& 40\left(1-\small\frac{1}{2}\normalsize\right)^{-6} \\[6pt] &=& 40\times \left(\small\frac{1}{2}\normalsize\right)^{-6} \\[6pt] &=& 40\times 2^6 \\[6pt] &=& 40\times 64 \\[6pt] &=& 2560 \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  Stationary points occur when the gradient is zero, i.e. when the derivative equals zero.

So our first steps are to differentiate and then fully factorise the derivative:

$$ \begin{eqnarray} f'(x) &=& 3x^2-3 \\[6pt] &=& 3(x^2-1) \\[6pt] &=& 3(x+1)(x-1) \\[6pt] \end{eqnarray} $$

Now, \(f'(x)=0\) at any stationary points, so

$$ \begin{gather} x+1=0\:\:\small\textsf{or}\normalsize\:\:x-1=0 \\[6pt] x=-1\:\:\small\textsf{or}\normalsize\:\:x=1 \end{gather} $$

(b)  A function is strictly increasing where its derivative is strictly greater than zero.

We know from part (a) that there are stationary points when \(x=-1\) and \(x=1\small.\)

One way of seeing where the derivative is positive is by using a nature table:

$$ \begin{array} {|c|c|} \hline x & \rightarrow & -1 & \rightarrow & 1 & \rightarrow \\ \hline f'(x) & + & 0 & - & 0 & + \\ \hline \end{array} $$

So \(f\) is strictly increasing when \(x \lt -1\) or \(x \gt 1.\)

Video solution by Clelland Maths 

This is an example of "optimisation". First we set some variables:

• Let \(x\) cm be the length.
• So the breadth is also \(x\) cm.
• Let \(h\) cm represent the height.
• Let \(A\) cm² be the surface area.

The surface area consists of the base and top (each of area \(x^2\)) and four sides (each of area \(xh\)).

So \(A = 2x^2 + 4xh\)

Notice that this formula for \(A\) involves two variables: \(x\) and \(h\). We need to eliminate one of them so that we can differentiate with respect to the other. In order to do this, we use the formula for the other piece of information that we are given: that the volume is \(125\) cm³.>/span>

So \(x^2 h=125\small,\) which gives us \(h=\large\frac{125}{x^2}\small.\)

Substitute this into the formula for the surface area and prepare it for differentiation:

$$ \begin{eqnarray} A &=& 2x^2 + 4xh \\[6pt] &=& 2x^2 + 4x\small\left(\frac{125}{x^2}\right)\normalsize \\[6pt] &=& 2x^2 + \small\frac{500}{x}\normalsize \\[6pt] &=& 2x^2 + 500x^{-1} \\[6pt] \end{eqnarray} $$

Now we differentiate with respect to \(x\):

$$ \frac{dA}{dx} = 4x-500x^{-2} $$

Stationary points occur when the derivative equals zero:

$$ \begin{eqnarray} 4x-500x^{-2} &=& 0 \\[6pt] \end{eqnarray} $$

Multiply through by \(x^2\) to get rid of the negative power:

$$ \begin{gather} 4x^3-500=0\\[6pt] 4x^3=500\\[6pt] x^3=125\\[6pt] x=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{125} \\[6pt] x=5 \end{gather} $$

We have now found a "critical value" of \(x\!=\!5\). However, we need to test that \(x\!=\!5\) really does give the minimum surface area. For that, we use a nature table:

$$ \begin{array} {|c|c|} \hline x & \rightarrow & 5 & \rightarrow \\ \hline \large\frac{dA}{dx}\normalsize & - & 0 & + \\ \hline \end{array} $$

This gives the shape of a minimum turning point, so \(x\!=\!5\) really does minimise the surface area.

All that remains is to find the height \(h,\) as the question asked for all the dimensions of the cuboid.

$$ h=\small\frac{125}{x^2}\normalsize = \small\frac{125}{5^2}\normalsize = 5 $$

So, unsurprisingly, our cuboid with minimum surface area is actually a \(5\!\times\!5\!\times\!5\) cm cube.

Video solution by Clelland Maths 

It is possible that the maximum or minimum values of the function are at the start or end of the interval. So we find the value of the function at the lower bound:

$$ \begin{eqnarray} f(-1) &=& 4(-1)^3+9(-1)^2-12(-1)+1 \\[6pt] &=& 4(-1)+9(1)+12+1 \\[6pt] &=& -\!4+9+12+1 \\[6pt] &=& 18 \\[6pt] \end{eqnarray} $$

Then we find the value at the upper bound:

$$ \begin{eqnarray} f(2) &=& 4(2^3)+9(2^2)-12(2)+1 \\[6pt] &=& 4(8)+9(4)-24+1 \\[6pt] &=& 32+36-24+1 \\[6pt] &=& 45 \\[6pt] \end{eqnarray} $$

If there are any maximum or minimum turning points within the interval, these might be where the maximum or minimum values are. So we need to differentiate and find any stationary points in the interval:

At stationary points, \(f'(x)=0\) so we solve:

Note that \(x=-2\) is outside the interval \(-1\leqslant x\leqslant 2\) so we disregard it.

However, we must find the value of \(f\left(\frac12\right)\) as it might be the maximum or minimum value within the interval.

We now compare the three values to see which is the maximum and which is the minimum:

\(f(-1)=18\), \(f(2)=45\) and \(f\left(\frac{1}{2}\right)=-2\frac{1}{4}\small.\)

So, in the given interval, the minimum value of \(f\) is \(-2\frac{1}{4}\) and the maximum value is \(45\).

Video solution by Clelland Maths 

This 2-mark question is a simple application of the chain rule.

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& 12(4x-1)^{11}(4) \\[6pt] &=& 48(4x-1)^{11} \end{eqnarray} $$

Video solution by Clelland Maths 

First we prepare \(d(t)\) for differentiation:

$$ \begin{eqnarray} d(t) &=& \small\frac{1}{2t} \\[6pt] &=& \small\frac{1}{2}\,\normalsize t^{-1} \end{eqnarray} $$

Now we differentiate:

$$ \begin{eqnarray} d'(t) &=& -\!\small\frac{1}{2}\,\normalsize t^{-2} \\[6pt] &=& -\!\frac{1}{2t^2} \end{eqnarray} $$

Finally, we substitute \(t=5\):

$$ \begin{eqnarray} d'(5) &=& -\!\frac{1}{2(5^2)}\\[6pt] &=& -\!\frac{1}{2(25)}\\[6pt] &=& -\!\frac{1}{50} \end{eqnarray} $$

Video solution by Clelland Maths 

(a)  To find the \(x\)-coordinate of the stationary point, we first differentiate:

$$ \begin{eqnarray} y &=& 6x-2\sqrt{\!x^3} \\[6pt] &=& 6x-2x^{\large\frac{3}{2}\normalsize} \\[12pt] \frac{dy}{dy} &=& 6-3x^{\large\frac12\normalsize} \\[6pt] &=& 6-3\sqrt{x} \\[6pt] \end{eqnarray} $$

Now we find the stationary point:

$$ \begin{gather} \frac{dy}{dy}=0\\[6pt] 6-3\sqrt{x}=0\\[6pt] 3\sqrt{x}=6\\[6pt] \sqrt{x}=2\\[6pt] x=4\\[6pt] \end{gather} $$

(b)  The \(x\)-coordinate of the stationary point is within the interval \(1\!\leqslant\!x\!\leqslant\!9\small,\) so we have to evaluate \(y\) when \(x=1\small,\) \(x=4\) and \(x=9\small.\)

\(x=1\implies y=6\!\times\!1-2\sqrt{\!1^3}=6-2=4\)

\(x=4\implies y=6\!\times\!4-2\sqrt{\!4^3}=24-16=8\)

\(x=9\implies y=6\!\times\!9-2\sqrt{\!9^3}=54-54=0\)

So the least value of \(y\) in the given interval is \(0\small,\) and its greatest value is \(8\small.\)

Video solution by Clelland Maths 

The process here is to find the derivative \(f'(x)\) and substitute \(x=2\) into it.

If \(f'(2)\) is positive, \(f\) is increasing when \(x=2\small.\) If negative, \(f\) is decreasing.

$$ \begin{eqnarray} f'(x) &=& 3x^2-7 \\[6pt] \therefore f'(2) &=& 3(2^2)-7 \\[6pt] &=& 12-7 \\[6pt] &=& 5 \\[6pt] \end{eqnarray} $$

\(f'(2)\gt 0\) so \(f\) is increasing when \(x=2\small.\)

Video solution by Clelland Maths 

First we differentiate, using the chain rule:

$$ \begin{eqnarray} f'(x) &=& 4\times 3\tiny\ \normalsize \text{cos}\left(3x-\small\frac{\pi}{3}\normalsize\right) \\[6pt] &=& 12\,\text{cos}\left(3x-\small\frac{\pi}{3}\normalsize\right) \\[6pt] \end{eqnarray} $$

Then we substitute \(x=\large\frac{\pi}{6}\small.\)

$$ \begin{eqnarray} f'\small\left(\frac{\pi}{6}\right)\normalsize &=& 12\,\text{cos}\left(\small\frac{3\pi}{6}-\small\frac{\pi}{3}\normalsize\right) \\[9pt] &=& 12\,\text{cos}\left(\small\frac{3\pi}{6}\normalsize-\small\frac{2\pi}{6}\normalsize\right) \\[9pt] &=& 12\,\text{cos}\,\small\frac{\pi}{6}\normalsize \\[9pt] &=& 12\,\small\left(\frac{\sqrt{3}}{2}\right)\normalsize \\[9pt] &=& 6\sqrt{3} \end{eqnarray} $$

Video solution by Clelland Maths 

Before differentiating, we need to express the second term in the form \(ax^n.\)

$$ \begin{eqnarray} y &=& x^{\large\frac{5}{3}\normalsize}-\small\frac{10}{x^4}\normalsize \\[6pt] &=& x^{\large\frac{5}{3}\normalsize}-10x^{-4}\normalsize \end{eqnarray} $$

Now that \(y\) is in the appropriate form, we are ready to differentiate:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& \small\frac{5}{3}\normalsize x^{\large\frac{2}{3}\normalsize}+40x^{-5} \end{eqnarray} $$

There is no need to go any further than the line above, as the question doesn't specify the form of the answer, but if you prefer, it could be expressed as follows:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& \small\frac{5}{3}\,\normalsize\sqrt[\leftroot{-1}\uproot{5}\scriptstyle 3]{x^2}+\small\frac{40}{x^5}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths 

We differentiate and then fully factorise the derivative:

$$ \begin{eqnarray} f'(x) &=& 6x^2+18x-24 \\[6pt] &=& 6(x^2+3x-4) \\[6pt] &=& 6(x+4)(x-1) \\[6pt] \end{eqnarray} $$

\(f'(x)=0\) at stationary points, so

$$ \begin{gather} x+4=0\:\:\small\textsf{or}\normalsize\:\:x-1=0 \\[6pt] x=-4\:\:\small\textsf{or}\normalsize\:\:x=1 \end{gather} $$

To determine the interval in which the function is strictly decreasing, we can use a nature table:

$$ \begin{array} {|c|c|} \hline x & \rightarrow & -4 & \rightarrow & 1 & \rightarrow \\ \hline f'(x) & + & 0 & - & 0 & + \\ \hline \end{array} $$

So \(f\) is strictly decreasing when \(-4\lt x\lt 1\small.\)

Video solution by Clelland Maths 

This 2-mark question is a very simple application of the chain rule.

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& 7(5x^2+3)^{6}(10x) \\[6pt] &=& 70x(5x^2+3)^6 \end{eqnarray} $$

Video solution by Clelland Maths 

First we prepare \(f(x)\) for differentiation:

$$ \begin{eqnarray} f(x) &=& 12\,\sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x} \\[6pt] &=& 12x^{\large\frac13} \end{eqnarray} $$

Now we differentiate:

$$ \begin{eqnarray} f'(x) &=& 4x^{-\large\frac23} \\[6pt] \end{eqnarray} $$

Finally, we use the fact that \(f'(a)=1\) to find \(a\small.\)

$$ \begin{eqnarray} 4a^{-\large\frac23\normalsize} &=& 1 \\[6pt] a^{-\large\frac23\normalsize} &=& \small\frac14 \\[6pt] a^{\large\frac23\normalsize} &=& 4\\[6pt] \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{a^2} &=& 4\\[6pt] a^2 &=& 4^3\\[6pt] a^2 &=& 64\\[6pt] a &=& \sqrt{64}\\[6pt] a &=& 8 \end{eqnarray} $$

Note that we are told in the question that \(x\!\gt\!0\) so we shouldn't give \(a=-8\) as a second solution.

Video solution by Clelland Maths 

First we prepare the function for differentiation:

$$ y = 8x^{-3}$$

Then we differentiate:

$$ \small\frac{dy}{dx}\normalsize = -24x^{-4} $$

Now we substitute \(x=2\) to find the gradient of the tangent:

$$ \begin{eqnarray} m_{\textsf{tan}} &=& -\!24(2^{-4}) \\[6pt] &=& -\!\frac{24}{2^4} \\[6pt] &=& -\!\frac{24}{16} \\[6pt] &=& -\!\frac{3}{2} \end{eqnarray} $$

Before we can find the equation of the tangent, we need to know the \(y\)-coordinate of the point of contact:

$$ y=\frac{8}{x^3}=\frac{8}{2^3}=\frac{8}{8}=1 $$

Finally, we use the Nat 5 method to find the equation of the straight line through \((2,1)\) with gradient \(-\large\frac32\small.\)

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-1 &=& -\!\small\frac32\normalsize(x-2)\\[6pt] 2(y-1) &=& -\!3(x-2)\\[6pt] 2y-2 &=& -\!3x+6\\[6pt] 2y &=& -\!3x+8\\[6pt] \end{eqnarray} $$

Note that the equation of the tangent could be given in various other forms, including \(3x+2y=8\) or \(y=-\large\frac32\normalsize x+4\small.\)

First we differentiate:

$$ \small\frac{dy}{dx}\normalsize = 3x^2-4x $$

Now we substitute \(x=2\) to find the gradient of the tangent:

$$ \begin{eqnarray} m_{\textsf{tan}} &=& 3(2^2)-4(2) \\[6pt] &=& 3(4)-8 \\[6pt] &=& 12-8 \\[6pt] &=& 4 \end{eqnarray} $$

Before we can find the equation of the tangent, we need to know the \(y\)-coordinate of the point of contact:

$$ \begin{eqnarray} y &=& x^3-2x^2+5 \\[6pt] &=& 2^3-2(2^2)+5 \\[6pt] &=& 8-8+5 \\[6pt] &=& 5 \end{eqnarray} $$

Finally, we use the Nat 5 method to find the equation of the straight line through \((2,\,5)\) with gradient \(4\):

$$ \begin{eqnarray} y-b &=& m(x-a)\\[6pt] y-5 &=& 4(x-2)\\[6pt] y-5 &=& 4x-8\\[6pt] y &=& 4x-3 \end{eqnarray} $$