Differentiation

The six derivatives, and the two you are not given

Cambridge lists exactly six functions whose derivatives you must know for this component: $\frac{d}{dx}(e^x)=e^x$, $\frac{d}{dx}(\ln x)=\frac{1}{x}$, $\frac{d}{dx}(\sin x)=\cos x$, $\frac{d}{dx}(\cos x)=-\sin x$, $\frac{d}{dx}(\tan x)={\sec }^{2}x$ and $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$. Together with the powers you met in Pure 1, that is the whole toolkit.

Two of the inverse trigonometric functions are deliberately left out. The syllabus states that the derivatives of ${\sin }^{-1}x$ and ${\cos }^{-1}x$ are not required, so do not learn them and do not expect a question on them. ${\tan }^{-1}x$ is the exception, and it is worth remembering that its derivative is not a trigonometric function at all: it is the rational function $\frac{1}{1+x^2}$.

Constant multiples, sums and differences work term by term, exactly as they did for polynomials: $\frac{d}{dx}(3e^x-2\ln x)=3e^x-\frac{2}{x}$, and $\frac{d}{dx}(4\sin x+5\cos x)=4\cos x-5\sin x$. The only sign to watch is the one in front of the cosine: the derivative of $\cos$ carries a minus, so a minus in front of $\cos$ becomes a plus after differentiating.

Composite functions and the chain rule

A composite function is one function inside another, such as $e^{3x^2-1}$ or $\ln (2x+1)$. The chain rule is $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$, which in words is: differentiate the outside function, leave the inside alone, then multiply by the derivative of the inside. Write $u$ for the inside when the chain is long enough that you might lose track.

Four patterns cover almost every question. For a power, $\frac{d}{dx}(2x+1{)}^5=5(2x+1{)}^4\times 2=10(2x+1{)}^4$. For an exponential, $\frac{d}{dx}(e^{3x^2-1})=e^{3x^2-1}\times 6x=6x{e}^{3x^2-1}$. For a logarithm, $\frac{d}{dx}(\ln (2x+1))=\frac{1}{2x+1}\times 2=\frac{2}{2x+1}$. For a trigonometric function, $\frac{d}{dx}(\sin 3x)=3\cos 3x$ and $\frac{d}{dx}(\tan 2x)=2{\sec }^{2}2x$.

The most common lost mark in this topic is a missing inner derivative. $\frac{d}{dx}(e^{3x})$ is $3e^{3x}$, not $e^{3x}$, and $\frac{d}{dx}(\ln 5x)$ is $\frac{1}{x}$, because $\ln 5x=\ln 5+\ln x$ and the constant $\ln 5$ differentiates away. Whenever you differentiate a composite, ask yourself what the inside would contribute.

Products: the product rule

When two functions of $x$ are multiplied, the product rule is $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Write down $u$, $v$, $\frac{du}{dx}$ and $\frac{dv}{dx}$ on four separate lines before you substitute anything; the algebra is then mechanical, and the examiner can see which marks you have earned.

For $y=x^2\ln x$, take $u=x^2$ and $v=\ln x$, so $\frac{du}{dx}=2x$ and $\frac{dv}{dx}=\frac{1}{x}$. Then $\frac{dy}{dx}=x^2\times \frac{1}{x}+\ln x\times 2x=x+2x\ln x$, which is also written $x(1+2\ln x)$. The first term collapses because $x^2\times \frac{1}{x}=x$, and that simplification is often a mark of its own.

For $y=x{e}^{1-x^2}$, take $u=x$ and $v=e^{1-x^2}$. Here $\frac{dv}{dx}$ needs the chain rule: $\frac{dv}{dx}=-2x{e}^{1-x^2}$. So $\frac{dy}{dx}=x(-2x{e}^{1-x^2})+e^{1-x^2}=(1-2x^2)e^{1-x^2}$, and taking the common factor $e^{1-x^2}$ out at the end is what turns a two-term answer into the single neat product Cambridge expects.

• The order of the two terms does not matter, but both must be there
• The product rule applies to $x\sin x$, $x^2{e}^x$, $\ln x\cos x$ and any other product of two functions of $x$
• If one factor is a constant, the rule is not needed: $\frac{d}{dx}(5\ln x)=\frac{5}{x}$
• If one factor is a power of $x$, check whether multiplying out is quicker: $x^2\times x^3$ is $x^5$ before you differentiate

Quotients: the quotient rule

When one function is divided by another, the quotient rule is $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$. Read the numerator in order: bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared. Reversing those two terms gives the negative of the correct answer, so the order is not cosmetic.

For $y=\frac{x+1}{2x-3}$, take $u=x+1$ and $v=2x-3$, so $\frac{du}{dx}=1$ and $\frac{dv}{dx}=2$. Then $\frac{dy}{dx}=\frac{(2x-3)\times 1-(x+1)\times 2}{(2x-3{)}^2}=\frac{2x-3-2x-2}{(2x-3{)}^2}=\frac{-5}{(2x-3{)}^2}$. Notice how much cancels: the $2x$ terms disappear, which is why the answer is so simple.

For $y=\frac{\ln x}{x}$, take $u=\ln x$ and $v=x$, giving $\frac{du}{dx}=\frac{1}{x}$ and $\frac{dv}{dx}=1$. Then $\frac{dy}{dx}=\frac{x\times \frac{1}{x}-\ln x\times 1}{x^2}=\frac{1-\ln x}{x^2}$, because $x\times \frac{1}{x}=1$. The quotient rule is also the quickest way to prove that $\frac{d}{dx}(\tan x)={\sec }^{2}x$, by writing $\tan x=\frac{\sin x}{\cos x}$.

Choosing the right rule

Read the expression's structure before you start. If the pieces are added or subtracted, differentiate term by term. If there is a function inside another function, you need the chain rule. If two functions are multiplied, you need the product rule. If one function is divided by another, you need the quotient rule - unless rewriting as a product of powers makes it easier.

Most exam expressions need two rules at once, and the trick is to decide the order. In $x^2\ln (3x)$, the product rule is the outer structure and the chain rule appears inside $\frac{dv}{dx}$. In $\frac{e^{2x}}{x}$, the quotient rule is the outer structure and the chain rule appears inside $\frac{du}{dx}$. Write the outer rule down first with $u$, $v$ and their derivatives as blanks, then fill the blanks using the chain rule.

One shortcut is worth knowing: a product rule where one factor is an exponential or a logarithm is usually the place to start, because $e^{f(x)}$ and $\ln f(x)$ both differentiate to something containing the original function, which then factors out at the end.

Parametric curves: finding the gradient from t

A curve given parametrically has $x$ and $y$ both written in terms of a third variable $t$, the parameter. The gradient is still $\frac{dy}{dx}$, but neither derivative is available directly, so you find both with respect to $t$ and divide: $\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}$. Both derivatives go through the chain rule, because every term contains $t$.

Take the syllabus's own example, $x=t-e^{2t}$ and $y=t+e^{2t}$. Differentiating gives $\frac{dx}{dt}=1-2e^{2t}$ and $\frac{dy}{dt}=1+2e^{2t}$, so $\frac{dy}{dx}=\frac{1+2e^{2t}}{1-2e^{2t}}$. At $t=0$ both exponentials are 1, so the gradient is $\frac{1+2}{1-2}=-3$. The answer is a number, and the question that asks for 'the gradient at the point where $t=0$' wants exactly that number.

If a question asks for the tangent, you need the point as well as the gradient. Substitute $t$ into the expressions for $x$ and $y$ to get the coordinates - for the curve above at $t=0$, $x=0-1=-1$ and $y=0+1=1$ - then use $y-y_1=m(x-x_1)$. Only the first derivative is required in this component: you are not asked for $\frac{d^{2}y}{d{x}^2}$ of a parametric curve.

• Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately, then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$
• Never write $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dx}{dt}$, and never write $\frac{dy}{dx}=\frac{dx}{dt}÷\frac{dy}{dt}$
• The gradient at a point is found by substituting the value of $t$, not the value of $x$
• If $\frac{dx}{dt}=0$ at the point, the tangent is vertical and $\frac{dy}{dx}$ is undefined there

Implicit differentiation

Some curves are not given as $y=f(x)$ at all, but as an equation linking $x$ and $y$, such as the syllabus's own example $x^2+y^2=xy+7$. To differentiate, treat $y$ as a function of $x$ and differentiate both sides with respect to $x$, term by term. Every term containing $y$ produces a factor of $\frac{dy}{dx}$, because of the chain rule: $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$, $\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}$, and $\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}$.

A product such as $xy$ needs the product rule as well, with $u=x$ and $v=y$: $\frac{d}{dx}(xy)=x\frac{dy}{dx}+y$. That single line is the step students most often get wrong, usually by writing $x\frac{dy}{dx}$ and forgetting the $y$.

Apply it to $x^2+y^2=xy+7$. Differentiating both sides gives $2x+2y\frac{dy}{dx}=y+x\frac{dy}{dx}$. Collect the $\frac{dy}{dx}$ terms on one side: $2y\frac{dy}{dx}-x\frac{dy}{dx}=y-2x$, so $\frac{dy}{dx}(2y-x)=y-2x$ and $\frac{dy}{dx}=\frac{y-2x}{2y-x}$. The answer is in terms of both $x$ and $y$, and that is normal: to find a gradient you substitute both coordinates of the point.

• Differentiate every term with respect to $x$, remembering that $y$ is a function of $x$
• A $y$ term differentiates to its ordinary derivative times $\frac{dy}{dx}$
• An $xy$ term needs the product rule: $x\frac{dy}{dx}+y$
• Collect, factorise out $\frac{dy}{dx}$, then divide - do not try to move terms one at a time
• Substitute both coordinates to get a numerical gradient, and check the point really does satisfy the original equation

Tangents and normals

Three of the four marks in a typical tangents and normals question are for getting the derivative and the point right, and only the last one is for the line itself. Find the gradient by differentiating and substituting the value given - $x$ for an explicit curve, $t$ for a parametric one, both $x$ and $y$ for an implicit one - and find the point by substituting into the equation of the curve (or into the parametric expressions for $x$ and $y$).

The tangent at a point with gradient $m$ is $y-y_1=m(x-x_1)$. The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{m}$, and its equation is $y-y_1=-\frac{1}{m}(x-x_1)$. If the question asks for the answer in a particular form, such as $ax+by+c=0$, rearrange at the very end and keep the coefficients whole numbers.

For $y=\ln (2x+1)$ at the point where $x=0$: the derivative is $\frac{2}{2x+1}$, which is 2 at $x=0$, and the point is $(0,\ln 1)=(0,0)$. The tangent is $y=2x$. The normal has gradient $-\frac{1}{2}$ and passes through the same point, so it is $y=-\frac{x}{2}$. One special case: if $m=0$ the tangent is horizontal and the normal is the vertical line $x=x_1$.

Logarithms and exponentials: simplify before you differentiate

The two composite derivatives $\frac{f^{\prime }(x)}{f(x)}$ for $\ln f(x)$ and $f^{\prime }(x)e^{f(x)}$ for $e^{f(x)}$ do most of the work in this section. $\ln (x^3+1)$ differentiates to $\frac{3x^2}{x^3+1}$, and $e^{\cos x}$ differentiates to $-\sin x{e}^{\cos x}$. In both cases the inside derivative is the only new ingredient.

The logarithm laws are worth using first. $\ln (x^3)=3\ln x$, so its derivative is $\frac{3}{x}$; differentiating $\ln (x^3)$ with the chain rule gives $\frac{3x^2}{x^3}=\frac{3}{x}$, which is the same answer by a longer route. $\ln \sqrt{x}=\frac{1}{2}\ln x$, so its derivative is $\frac{1}{2x}$. $\ln (x^2{e}^x)=2\ln x+x$, so its derivative is $\frac{2}{x}+1$ - and here the split is much easier than the product rule.

One further rewriting trick: any exponential with a different base can be written in terms of $e$. Since $a^x=e^{x\ln a}$, its derivative is $\ln a\times a^x$. This is not a new rule, just the chain rule applied to a base you have rewritten, and it is a useful escape if a question uses $2^x$ rather than $e^x$.

• $\ln (ab)=\ln a+\ln b$ turns a product inside a logarithm into a sum you can differentiate term by term
• $\ln (a^n)=n\ln a$ turns a power inside a logarithm into a multiple
• $e^{f(x)}$ is never zero, so a factor of it can always be taken out of an answer
• A constant inside a logarithm disappears when you differentiate: $\ln (7x)$ and $\ln x$ have the same derivative