This is a simple two-mark product rule question, in which neither of the terms requires the chain rule.

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

$$ \begin{matrix} u=x^7 \:&\: v=tan\,x \\ u'=7x^6 \:&\: v'=sec^{2}\,x \\ \end{matrix} $$

Now we apply the product rule:

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

This harder example requires both the product rule and the chain rule.

$$ \begin{matrix} u=e^{sin\,x} \:&\: v=sec\,x \\ u'=e^{sin\,x}cos\,x \:&\: v'=sec\,x\,tan\,x \\ \end{matrix} $$

Now we apply the product rule and simplify:

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

The simplification occurred because \(sec\,x=\large\frac{1}{cos\,x},\) so \(cos\,x\,sec\,x=1.\) You need to have your wits about you when working with the reciprocal trig functions!

The factorisation at the end is a matter of style preference. The second last line is also a perfectly acceptable final answer.

Both \(ln\,3x\) and \(cos^{-1}\,2x\) require the chain rule.

$$ \begin{matrix} u=ln\,3x \:&\: v=cos^{-1}\,2x \\ u'=\large\frac{3}{3x}\normalsize \:&\: v'=-\large\frac{2}{\sqrt{1-(2x)^2}}\normalsize \\ \phantom{u'}=\large\frac{1}{x}\normalsize \:&\: \phantom{v'}=-\large\frac{2}{\sqrt{1-4x^2}}\normalsize \\ \end{matrix} $$

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

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

This is a simple example of the quotient rule.

$$ \begin{matrix} u=2x-1 \:&\: v=1-x^2 \\ u'=2 \:&\: v'=-2x \\ \end{matrix} $$

Now we apply the quotient rule and simplify the numerator.

$$ \begin{eqnarray} f'(x) &=& \small\frac{u'\,v-u\,v'}{v^2} \\[8pt] &=& \small\frac{2(1-x^2)-(2x-1)(-2x)}{(1-x^2)^2}\\[8pt] &=& \small\frac{2(1-x^2)+2x(2x-1)}{(1-x^2)^2}\\[8pt] &=& \small\frac{2-2x^2+4x^2-2x}{(1-x^2)^2}\\[8pt] &=& \small\frac{2x^2-2x+2}{(1-x^2)^2} \end{eqnarray} $$

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

$$ \begin{matrix} u=e^{1+x^2} \:&\: v=1+x^2 \\ u'=2x\,e^{1+x^2} \:&\: v'=2x \\ \end{matrix} $$

Now we apply the quotient rule and simplify the numerator.

$$ \begin{eqnarray} f'(x) &=& \frac{u'\,v-u\,v'}{v^2} \\[8pt] &=& \frac{2x\,e^{1+x^2}(1+x^2)-e^{1+x^2}(2x)}{(1+x^2)^2}\\[8pt] &=& \frac{2x\large[\normalsize e^{1+x^2}(1+x^2)-e^{1+x^2}\large]\normalsize}{(1+x^2)^2}\\[8pt] &=& \frac{2x(e^{1+x^2}+x^{2}\,e^{1+x^2}-e^{1+x^2})}{(1+x^2)^2}\\[8pt] &=& \frac{2x(x^{2}\,e^{1+x^2})}{(1+x^2)^2}\\[8pt] &=& \frac{2x^{3}\,e^{1+x^2}}{(1+x^2)^2}\\[8pt] \end{eqnarray} $$

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

$$ \begin{eqnarray} \frac{dy}{dx} &=& \frac{1}{cosec\,x^2}\,.\,\frac{d}{dx}(cosec\,x^2)\\[6pt] &=& \frac{1}{cosec\,x^2}\,.-cosec\,x^2\,cot\,x^2.\frac{d}{dx}(x^2)\\[6pt] &=& -cot\,x^2\,.\, \frac{d}{dx}(x^2)\\[6pt] &=& -2x\,cot\,x^2 \end{eqnarray} $$

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

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

Now let's deal with the quotient separately:

$$ \begin{matrix} u=x \:&\: v=x^3-4 \\ u'=1 \:&\: v'=3x^2 \\ \end{matrix} $$

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

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 usually, but not always, say "use implicit differentiation" or tell you that a function "is defined implicitly." Nonetheless, if you are faced with an inseparable mixture of \(x\) and \(y,\) you should know to differentiate implicitly.

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

The key here is to understand that \(y\) is a function of \(x,\) not a constant, so it behaves as such in the product, quotient or chain rules.

This example requires the product rule.

$$ \begin{matrix} u=y \:&\: v=cot\,x \\ u'=\frac{dy}{dx} \:&\: v'=-\!cosec^{2}\,x \\ \end{matrix} $$

Now we apply the product rule to the entire equation:

$$\frac{dy}{dx}.cot\,x+y(-\!cosec^{2}\,x)-3y^2.\frac{dy}{dx}=2 $$

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

$$\frac{dy}{dx}(cot\,x-3y^2)=2+y\,cosec^{2}\,x $$

Finally, divide:

$$\frac{dy}{dx}=\frac{2+y\,cosec^{2}\,x}{cot\,x-3y^2} $$

For the quotient on the left:

$$ \begin{matrix} u=x \:&\: v=y \\ u'=1 \:&\: v'=\large\frac{dy}{dx}\normalsize \\ \end{matrix} $$

Differentiating implicitly:

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

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

Differentiating implicitly:

\begin{gather} \frac{y-x\large\frac{dy}{dx}\normalsize}{y^2} = \small\frac{dy}{dx}\\[10pt] y-x\small\frac{dy}{dx}\normalsize=y^{2}\small\frac{dy}{dx}\normalsize\\[10pt] y=(y^{2}+x)\small\frac{dy}{dx}\normalsize\\[10pt] \frac{dy}{dx}\normalsize=\frac{y}{y^2+x}\\[10pt] \end{gather}

The second derivative requires the quotient rule:

$$ \begin{matrix} u=y \:&\: v=y^2+x \\ u'=\large\frac{dy}{dx}\normalsize \:&\: v'=2y\,\large\frac{dy}{dx}\normalsize+1 \\ \end{matrix} $$

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

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

Now we can apply the chain rule:

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

The method for the second derivative is as follows:

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

This is an example of instantaneous speed.

$$ \begin{matrix} x=2t \:&\: y=sin\,t \\[8pt] \Large\frac{dx}{dt}\normalsize=2 \:&\: \Large\frac{dy}{dt}\normalsize=cos\,t \\[4pt] \end{matrix} $$

We do not need to find \(\Large\frac{dy}{dx}\normalsize\) 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.

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

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

Note: If you got \(2.24\) instead of \(2.04,\) set your calculator to radians, not degrees. Calculus always uses radians.

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

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

$$ \begin{gather} y = x^{x^{2}-2} \\[8pt] ln\,y = ln\,x^{x^{2}-2} \\[8pt] ln\,y = (x^2\!-\!2)\,ln\,x \\[8pt] \end{gather} $$

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

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

This question may look 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.

$$ \begin{gather} e^{y}=\small\frac{(2x-1)e^{3x}}{(4x+1)^2}\normalsize \\[8pt] ln\,e^{y}=ln\,\small\frac{(2x-1)e^{3x}}{(4x+1)^2}\normalsize \\[8pt] y=ln\,\vert 2x\!-\!1\vert+ln\,e^{3x}-ln\,\vert 4x\!+\!1\vert^2 \\[8pt] y=ln\,\vert 2x\!-\!1\vert+3x-2\,ln\,\vert 4x\!+\!1\vert \\[8pt] y=ln\,(2x\!-\!1)+3x-2\,ln\,(4x\!+\!1) \\[8pt] \end{gather} $$

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

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:

$$ \small\frac{\textsf{dr}}{\textsf{dt}}\normalsize = \small\frac{\textsf{dr}}{\textsf{dV}}\normalsize\times \small\frac{\textsf{dV}}{\textsf{dt}}\normalsize $$

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

$$ \small\frac{\textsf{dV}}{\textsf{dr}}\normalsize = 4\pi r^2 $$

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

Note that the SQA will require units 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:

$$ \begin{matrix} u=\pi r^{2} \:&\: v=h \\ u'=2\pi r \large\frac{\textsf{dr}}{\textsf{dt}}\normalsize \:&\: v'=1.\large\frac{\textsf{dh}}{\textsf{dt}}\normalsize \\ =2\pi r (-0.02) \:&\: =1(0.01) \\ =-0.04\pi r \:&\: =0.01 \\ \end{matrix} $$

Putting this together:

$$ \begin{eqnarray} \small\frac{\textsf{dV}}{\textsf{dt}}\normalsize &=& u'\,v+u\,v' \\[6pt] &=& -\!0.04\pi rh + 0.01 \pi r^2 \\[6pt] \end{eqnarray} $$

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.