Advanced Higher Maths
Differentiation

This is a simple two-mark product rule question, as neither factor involves a composite function.

You may have been taught to lay out your working as follows, although with experience this isn't necessary:

Now we apply the product rule:

$$ \begin{eqnarray} f'(x) &=& u'\,v+u\,v' \\[6pt] &=& 7x^6\,\text{tan}\,x+x^{7}\,\text{sec}^{2}\,x \end{eqnarray} $$

This harder example requires both the product rule and the chain rule for the first factor.

Now we apply the product rule and simplify:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& u'\,v+u\,v' \\[6pt] &=& e^{\,\large{\text{sin}\,x}}\,\text{cos}\,x\,\text{sec}\,x+e^{\,\large\text{sin}\,x}\,\text{sec}\,x\,\text{tan}\,x \\[6pt] &=& e^{\,\large{\text{sin}\,x}}+e^{\,\large{\text{sin}\,x}}\,\text{sec}\,x\,\text{tan}\,x \\[6pt] &=& (1+\text{sec}\,x\,\text{tan}\,x)\,e^{\large{\,{\text{sin}}\,x}} \end{eqnarray} $$

Note 1: The \(\text{cos}\,x\,\text{sec}\,x\) disappeared because \(\text{sec}\,x\) is the reciprocal of \(\text{cos}\,x\small,\) so \(\text{cos}\,x\,\text{sec}\,x=1\small.\) When working with the reciprocal trig functions, keep your eyes open for this kind of thing.

Note 2: The factorisation in the last line is just a matter of style preference. The second last line is also a perfectly acceptable final answer.

This is a product rule question in which both factors, \(\text{ln}\,3x\) and \(\text{cos}^{-1}\,2x\small,\) require the chain rule.

Now we apply the product rule and simplify as much as possible – which in this case isn't much!

$$ \begin{eqnarray} f'(x) &=& u'\,v+u\,v' \\[6pt] &=& \small\frac{1}{x}\normalsize\,\text{cos}^{-1}\,2x-\biggl(\text{ln}\,3x\biggr)\biggl(\small\frac{2}{\sqrt{1-4x^2}}\normalsize\biggr)\\[6pt] &=& \small\frac{\text{cos}^{-1}\,2x}{x}\normalsize-\small\frac{2\,\text{ln}\,3x}{\sqrt{1-4x^2}} \end{eqnarray} $$

This is a simple example of the quotient rule.

You may have been taught to lay out your working as follows, although with experience this isn't necessary:

Now we apply the quotient rule and simplify the numerator.

$$ \begin{eqnarray} f'(x) &=& \frac{u'\,v-u\,v'}{v^2} \\[10pt] &=& \frac{2(1-x^2)-(2x-1)(-2x)}{(1-x^2)^2}\\[10pt] &=& \frac{2(1-x^2)+2x(2x-1)}{(1-x^2)^2}\\[10pt] &=& \frac{2-2x^2+4x^2-2x}{(1-x^2)^2}\\[10pt] &=& \frac{2x^2-2x+2}{(1-x^2)^2} \end{eqnarray} $$

Note 1: We should not expand the denominator. Expansion is not simplification. Often, it's the reverse!

Note 2: If you prefer, you could write the numerator as \(2(x^2-x+1)\small.\) That's just a style preference.

This is a slightly harder example of the quotient rule. The numerator requires the chain rule.

Now we apply the quotient rule and simplify the numerator by factorising.

$$ \begin{eqnarray} f'(x) &=& \frac{u'\,v-u\,v'}{v^2} \\[12pt] &=& \frac{2x\,e^{1+x^2}(1+x^2)-e^{1+x^2}(2x)}{\left(1+x^2\right)^2}\\[12pt] &=& \frac{2x\ e^{1+x^2}\,\bigl\{(1+x^2)-1\bigr\}}{\left(1+x^2\right)^2}\\[12pt] &=& \frac{2x\,e^{1+x^2}(x^2)}{\left(1+x^2\right)^2}\\[12pt] &=& \frac{2x^{3}\,e^{1+x^2}}{\left(1+x^2\right)^2}\\[12pt] \end{eqnarray} $$

This is fairly straightforward double application of the chain rule with two of the standard derivatives.

$$ \begin{eqnarray} \frac{dy}{dx} &=& \frac{1}{\text{cosec}\,x^2}\,.\,\frac{d}{dx}(\text{cosec}\,x^2)\\[10pt] &=& \frac{1}{\text{cosec}\,x^2}\,.-\text{cosec}\,x^2\,\text{cot}\,x^2\,.\,\frac{d}{dx}(x^2)\\[10pt] &=& \bigl(-\text{cot}\,x^2\bigr)\bigl(2x\bigr)\\[10pt] &=& -\!2x\,\text{cot}\,x^2 \end{eqnarray} $$

This will need the quotient rule for the inner function within a chain rule, so it's going to get messy!

Note the method below to simplify the 'fraction in a fraction'. We multiply top and bottom by the denominator.

$$ \begin{eqnarray} f'(x) &=& \frac{1}{1+\left(\displaystyle\small\frac{x}{x^3-4}\right)^2}\ .\,\frac{d}{dx}\left(\frac{x}{x^3-4}\right)\\[12pt] &=& \frac{(x^3-4)^2}{(x^3-4)^2+x^2}\ .\,\frac{d}{dx}\left(\frac{x}{x^3-4}\right)\\[8pt] \end{eqnarray} $$

Now let's deal with the quotient rule separately:

$$ \begin{eqnarray} \frac{d}{dx}\left(\frac{x}{x^3-4}\right) &=& \frac{u'\,v-u\,v'}{v^2} \\[12pt] &=& \frac{(1)(x^3-4)-x(3x^2)}{(x^3-4)^2} \\[12pt] &=& \frac{x^3-4-3x^3}{(x^3-4)^2} \\[12pt] &=& \frac{-2x^3-4}{(x^3-4)^2} \\[8pt] \end{eqnarray} $$

Substituting this back into \(f'(x)\):

$$ \begin{eqnarray} f'(x) &=& \frac{(x^3-4)^2}{(x^3-4)^2+x^2}\ .\,\frac{-2x^3-4}{(x^3-4)^2}\\[8pt] &=& \frac{-2x^3-4}{(x^3-4)^2+x^2}\\[8pt] \end{eqnarray} $$

Finally, we substitute \(x\!=\!2.\)

$$ \begin{eqnarray} f'(2) &=& \frac{-2(2)^3-4}{(2^3-4)^2+2^2}\\[8pt] &=& \frac{-16-4}{4^2+4}\\[8pt] &=& \frac{-20}{20}\\[8pt] &=& -\!1 \end{eqnarray} $$

Advanced Higher exam papers don't always say "use implicit differentiation" or tell you that a function "is defined implicitly". If the function has an inseparable combination of \(x\) and \(y\small,\) you should know to differentiate implicitly.

The question won't always remind you that \(\large\frac{dy}{dx}\normalsize\) involves both \(x\) and \(y,\) but that is always the case.

Recall that \(x\) and \(y\) are variables, not constants, so they behave as such in the product, quotient and chain rules.

This example requires the product rule.

Now we apply the product rule to the entire equation:

Bring the \(\large\frac{dy}{dx}\normalsize\) terms to one side, and the other terms to the other side, so that we can factorise:

Finally, divide:

For the quotient on the left:

Differentiating implicitly, with respect to \(x\):

$$ \begin{gather} \frac{u'\,v-u\,v'}{v^2} = e^{\large{y}}\,.\,\small\frac{dy}{dx}\normalsize \\[10pt] \frac{y-x\large\frac{dy}{dx}\normalsize}{y^2} = e^{\large{y}}\,.\,\small\frac{dy}{dx}\normalsize\\[10pt] y-x\small\frac{dy}{dx}\normalsize = y^2\,e^{\large{y}}\,.\,\small\frac{dy}{dx}\normalsize\\[12pt] y = y^2\,e^{\large{y}}\,.\,\small\frac{dy}{dx}\normalsize+x\small\frac{dy}{dx}\normalsize\\[12pt] y = \small\frac{dy}{dx}\normalsize\left(y^2\,e^{\large{y}}+x\right)\\[12pt] \frac{dy}{dx} = \frac{y}{y^2\,e^{\large{y}}+x}\\[12pt] \end{gather} $$

Note 1: This isn't the only reasonable form for the final answer. For example, we could substitute \(\displaystyle\small\frac{x}{y}\) for \(e^y\) and simplify to \(\displaystyle\small\frac{y}{x(y+1)}\) if we think that looks 'prettier'. Such decisions are often just a matter of personal preference.

Note 2: Although the wording of this question notes that the function is defined implicitly, it does not instruct us to use implicit differentiation. We need implicit differentiation when the \(x\) and \(y\) variables are inseparable, as in example 8 above, but that isn't the case here, because the equation can be rearranged to \(x=y\,e^y\small.\) So we could use the product rule to find \(\displaystyle\small\frac{dx}{dy}\) and then take the reciprocal, giving \(\displaystyle\small\frac{dy}{dx}=\displaystyle\small\frac{1}{e^{y}(y+1)}\small.\) Why not try this for yourself?

So that we can move quickly to finding the second derivative, we have given this function the same left hand side as example 9 above.

Differentiating implicitly:

The second derivative requires the quotient rule:

$$ \begin{eqnarray} \frac{d^{2}y}{dx^2} &=& \frac{u'\,v-u\,v'}{v^2}\\[12pt] &=& \frac{\displaystyle\small\frac{dy}{dx}\normalsize\bigl(x+y^2\bigr)-y\,\bigl(1+2y\,\displaystyle\small\frac{dy}{dx}\normalsize\bigr)}{\left(x+y^2\right)^2}\\[12pt] &=& \frac{\large\frac{y(x\,+\,y^2)}{x\,+\,y^2}\normalsize-y\,\biggl(1+2y\left(\large\frac{y}{x\,+\,y^2}\normalsize\right)\biggr)}{\left(x+y^2\right)^2}\\[12pt] &=& \frac{y-y-\large\frac{2y^3}{x\,+\,y^2}\normalsize}{\left(x+y^2\right)^2}\\[12pt] &=& \frac{-\large\frac{2y^3}{x\,+\,y^2}\normalsize}{\left(x+y^2\right)^2}\\[12pt] &=& -\!\frac{2y^3}{\left(x+y^2\right)^3}\\[8pt] \end{eqnarray} $$

Note: As the original equation \(\displaystyle\small\frac{x}{y}\normalsize=y+1\) can, in fact, be expressed as \(x=y^2+y\small,\) implicit differentiation isn't actually needed to find the first and second derivatives. If the question hadn't demanded implicit differentiation, we could have found \(\displaystyle\small\frac{dx}{dy}\) and then used the reciprocal rule for \(\displaystyle\small\frac{dy}{dx}\small,\) yielding \(\displaystyle\small\frac{dy}{dx\normalsize}=\displaystyle\small\frac{1}{2y+1}\small.\) However, a similar approach wouldn't work for the second derivative. You would have to use the quotient rule and chain rule, giving \(\displaystyle\small\frac{d^{2}y}{dx^2}\normalsize=-\displaystyle\small\frac{2}{(2y+1)^3}\small.\) These expressions, in terms of only \(y\small,\) are equivalent to the derivatives obtained in the working above. Why not try it this way for yourself?

First we differentiate each of the component functions with respect to \(t\):

$$ \begin{matrix} x=(\text{ln}\,t)^2\: \:\:&\:\:\: y=2\,\text{ln}\,t \\[8pt] \displaystyle\frac{dx}{dt}\normalsize=\biggl(2\,\text{ln}\,t\biggr)\biggl(\displaystyle\frac{1}{t}\normalsize\biggr) \:\:\:&\:\:\: \displaystyle\frac{dy}{dt}\normalsize=\displaystyle\frac{2}{t}\normalsize\\[8pt] \phantom{.}=\displaystyle\frac{2\,\text{ln}\,t}{t}\normalsize \:\:\:&\:\:\: \phantom{ }\\[4pt] \end{matrix} $$

Then we apply the chain rule:

$$ \begin{eqnarray} \frac{dy}{dx} &=& \frac{dy}{dt}\,.\,\frac{dt}{dx}\,\\[8pt] &=& \frac{2}{t}\,.\,\frac{t}{2\,\text{ln}\,t}\\[8pt] &=& \frac{1}{\text{ln}\,t} \end{eqnarray} $$

The method for the second derivative is as follows:

$$ \begin{eqnarray} \frac{d^{2}y}{dx^2} &=& \frac{d}{dx}\left(\frac{dy}{dx}\right)\\[8pt] &=& \frac{d}{dt}\left(\frac{dy}{dx}\right)\,.\,\frac{dt}{dx}\\[8pt] &=& \frac{d}{dt}\left((\text{ln}\,t)^{-1}\right)\,.\,\frac{t}{2\,\text{ln}\,t}\\[8pt] &=& -\!(\text{ln}\,t)^{-2}\,\biggl(\frac{1}{t}\biggr)\,.\,\frac{t}{2\,\text{ln}\,t}\\[8pt] &=& -\!(\text{ln}\,t)^{-2}\,.\,\frac{1}{2\,\text{ln}\,t}\\[8pt] &=& -\!\frac{1}{2\,(\text{ln}\,t)^{3}} \end{eqnarray} $$

This is an example of instantaneous speed.

First we find the component derivatives, with respect to \(t\):

We do not need to find \(\displaystyle\small\frac{dy}{dx}\normalsize\) in this example. We can apply the formula for instantaneous speed, which is just the magnitude of the velocity, hence the obvious resemblance of the formula to Pythagoras' Theorem.

$$ \begin{eqnarray} \small\textsf{Speed }\ &=& \sqrt{\biggl(\displaystyle\frac{dx}{dt}\biggr)^2+\biggl(\displaystyle\frac{dy}{dt}\biggr)^2\ }\\[8pt] &=& \sqrt{4+\text{cos}^{2}\,t\ }\\[8pt] \end{eqnarray} $$

When \(t=2\small,\) the speed is \(\sqrt{4+\text{cos}^{2}\,2\ }\) \(\approx 2.04\) m/s.

Note: If you got \(2.24\) instead of \(2.04,\) set your calculator to radians, not degrees. Calculus always uses radians. For example, the derivative of \(\text{sin}\,t\) is only \(\text{cos}\,t\) when \(t\) is in radians.

Advanced Higher exam papers sometimes say "use logarithmic differentiation" or ask you to "differentiate logarithmically." However, the specification  says that you should be able to recognise when logarithmic differentiation is required. Usually it is signalled by having \(x\) in an exponent.

The usual method is to take natural logs of both sides, use the Higher log laws to express powers as products, and then to differentiate implicitly.

Now we use implicit differentiation. The right hand side also requires the product rule.

$$ \begin{eqnarray} \small\frac{1}{y}\normalsize\,.\,\small\frac{dy}{dx}\normalsize &=& 2x\,\text{ln}\,x+\biggl(x^{2}\!-\!2\biggr)\biggl(\small\frac{1}{x}\normalsize\biggr)\\[8pt] &=& 2x\,\text{ln}\,x+x-\small\frac{2}{x}\normalsize\\[8pt] \small\frac{dy}{dx}\normalsize &=& y\left(2x\,\text{ln}\,x+x-\small\frac{2}{x}\normalsize\right)\\[8pt] &=& x^{\large{x^{2}-2}}\small\left(\normalsize2x\,\text{ln}\,x+x-\small\frac{2}{x}\right)\normalsize \end{eqnarray} $$

This question may look a lot harder than the previous example, but it isn't really. After taking natural logarithms of both sides, the log laws will simplify it hugely, making the differentiation easier than you might expect.

Note that the modulus signs were unnecessary as the question told us that \(x\gt\frac{1}{2}\small.\)

Now we can differentiate:

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& \small\frac{2}{2x\!-\!1}\normalsize+3-2\small\left(\frac{4}{4x\!+\!1}\right)\normalsize\\[8pt] &=& \small\frac{2}{2x\!-\!1}\normalsize+3-\small\frac{8}{4x\!+\!1}\normalsize \\[8pt] \end{eqnarray} $$

This question involves related rates of change. We have been told the rate at which the volume is changing with respect to time: \(\large\frac{\textsf{dV}}{\textsf{dt}}\normalsize =20\) cm³ s^–1. We are being asked to find the value of \(\large\frac{\textsf{dr}}{\textsf{dt}}\normalsize\) at a specific time.

The key to these types of questions is to use the chain rule:

\(\large\frac{\textsf{dr}}{\textsf{dV}}\normalsize\) is the reciprocal of \(\large\frac{\textsf{dV}}{\textsf{dr}}\normalsize\) so we differentiate the volume formula:

So now we combine this information:

$$ \begin{eqnarray} \small\frac{\textsf{dr}}{\textsf{dt}}\normalsize &=& \small\frac{\textsf{dr}}{\textsf{dV}}\normalsize\times \small\frac{\textsf{dV}}{\textsf{dt}}\normalsize \\[8pt] &=& \small\frac{1}{4\pi r^2}\normalsize\times 20 \\[8pt] &=& \small\frac{5}{\pi r^2}\normalsize \\[8pt] \end{eqnarray} $$

When \(r\!=\!5\small,\) this is \(\large\frac{5}{\pi(5^2)}\normalsize=\large\frac{1}{5\pi}\normalsize\) cm s^–1.

Note that units of measurement are required in your final answer. The rate of change of a length in centimetres with respect to a time in seconds will of course be cm s^–1.

This is another related rates of change question, but unlike the previous example, this question involves two variables. Both the radius and height of the cylinder are changing.

So \(\large\frac{\textsf{dr}}{\textsf{dt}}\normalsize =-0.02\) (negative as it's decreasing) and \(\large\frac{\textsf{dh}}{\textsf{dt}}\normalsize =0.01\) (positive as it's increasing).

We are being asked to find \(\large\frac{\textsf{dV}}{\textsf{dt}}\normalsize\) with given values of \(r\) and \(h\) substituted into it.

It is important to understand that differentiating \(V\) requires the product rule, because both \(r\) and \(h\) are variables.

Differentating \(V\) with respect to \(t\) also requires implicit differentiation, as follows:

Putting this together:

$$ \begin{eqnarray} \small\frac{\textsf{dV}}{\textsf{dt}}\normalsize &=& u'\,v+u\,v' \\[4pt] &=& -\!0.04\pi rh + 0.01 \pi r^2 \\[10pt] \end{eqnarray} $$

So the rate of change of volume when \(r=0.6\) and \(h=2\) equals

$$ \begin{eqnarray} \, &\,& -\!0.04\pi(0.6)(2) + 0.01 \pi (0.6^2) \\[6pt] &=& -\!0.048\pi + 0.0036\pi\\[6pt] &=& -\!0.0444\pi \\[6pt] \end{eqnarray} $$

We must write this with the correct units, so the final answer is \(-0.0444\pi\) m³ s^–1.

To show that the point \((-1,\,5)\) lies on the curve, we need to find a single value of the parameter \(t\) that will give us both \(x\!=\!-1\) and \(y\!=\!5\small.\)

The equation for \(x\) is simpler than the equation for \(y\small,\) so let's start as follows:

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

Now we consider the other equation, \(y=2t^2-t+2\small,\,\) when \(y=5\small.\)

\(t\!=\!0\) would give us \(5\!=\!2\) so we can quickly eliminate that possibility!

Now we consider the other possibility: \(t\!=\!-1\small.\)

$$ \begin{gather} y=2t^2-t+2 \\[6pt] 5=2(-1)^2-(-1)+2\\[6pt] 5=2+1+2 \\[6pt] \end{gather} $$

This is true, so \(t=-1\) results in both \(x=-1\) and \(y=5\small,\) meaning that the point \((-1,\,5)\) lies on the curve.

Now we move to the second part of the question: finding the equation of the tangent through A\(\,(-1, 5)\small.\)

To do this, we need to differentiate to find the gradient of the tangent. So:

$$ \begin{matrix} x=t^2+t-1 \:&\: y=2t^2-t+2 \\[6pt] \displaystyle\small\frac{dx}{dt}\normalsize=2t+1 \:&\: \displaystyle\small\frac{dy}{dt}\normalsize=4t-1 \\[6pt] \end{matrix} $$

Applying the chain rule:

$$ \begin{eqnarray} \small\frac{dy}{dx} &=& \small\frac{dy}{dt}\normalsize\,.\small\frac{dt}{dx}\,\\[8pt] &=& \small\frac{4t-1}{2t+1}\\[8pt] \end{eqnarray} $$

Substituting \(t=-1\) to find the gradient:

$$ \begin{eqnarray} m &=& \small\frac{4(-1)-1}{2(-1)+1}\\[8pt] &=& \small\frac{-5}{-1}\\[8pt] &=& 5 \end{eqnarray} $$

Nearly done! All that remains is to find the equation of the tangent at A:

$$ \begin{gather} y-5=5\left(x-(-1)\right)\\[6pt] y-5=5(x+1)\\[6pt] y-5=5x+5\\[6pt] y=5x+10\\[6pt] \end{gather} $$