SQA Advanced Higher Mathematics

Exponential and natural logarithm functions

The exponential function $e^x$ is its own derivative, and the natural logarithm is the function whose derivative is the reciprocal:

Combined with the chain rule these give the two results used most often. For any differentiable $f(x)$,

So $\frac{d}{dx}\left(e^{3x^2 - x}\right) = (6x - 1)e^{3x^2 - x}$ and $\frac{d}{dx}\left(\ln(x^2 + 1)\right) = \frac{2x}{x^2 + 1}$.

The second result is often easier after the logarithm laws have been used to split the expression. Since $\ln\left(\dfrac{x^3}{\sqrt{x+1}}\right) = 3\ln x - \tfrac{1}{2}\ln(x+1)$, its derivative is $\dfrac{3}{x} - \dfrac{1}{2(x+1)}$, with no quotient rule needed.

The chain rule

If $y = f(u)$ and $u = g(x)$, then

Differentiate the outer function with the inner function left alone, then multiply by the derivative of the inner function. Functions built from up to three layers need the rule up to twice in one line.

For $y = \sin^3(2x)$ the layers are cube, sine and $2x$:

For $y = e^{\sin(x^2)}$ the layers are exponential, sine and square:

Writing each layer as a separate line of working, with $u$ and $v$ named, is slower but avoids dropping a factor. The most common slip is to forget the derivative of the innermost function.

The product rule

For $y = uv$, where $u$ and $v$ are functions of $x$,

To differentiate $y = x^3e^{-2x}$, take $u = x^3$ and $v = e^{-2x}$, so $u' = 3x^2$ and $v' = -2e^{-2x}$:

Factorising at the end is worth doing, because the next step is usually to set the derivative equal to zero or to evaluate it.

The quotient rule

For $y = \dfrac{u}{v}$,

To differentiate $y = \dfrac{\ln x}{x^2}$, take $u = \ln x$ and $v = x^2$:

The numerator order matters: it is $v\,u' - u\,v'$, and swapping the two terms changes the sign of the answer.

Derivatives of tan, cot, sec and cosec

These four are derived from sine and cosine with the quotient rule or the chain rule, and the derivations are part of what has to be known.

For $\tan x = \dfrac{\sin x}{\cos x}$:

For $\cot x = \dfrac{\cos x}{\sin x}$:

For $\sec x = (\cos x)^{-1}$, the chain rule gives

For $\operatorname{cosec} x = (\sin x)^{-1}$,

Both of the functions whose names begin with "co" have negative derivatives.

With a linear inner function, the chain rule gives $\frac{d}{dx}\left(\tan(ax)\right) = a\sec^2(ax)$, and in the same way for the others. For instance,

Combining the rules

Many questions need more than one rule in a single derivative. The usual order is to identify the outermost operation (a product, a quotient or a composite), apply that rule, and then differentiate each part, using the chain rule inside whichever part needs it.

For $y = x\operatorname{cosec}x$, the product rule gives

For $y = e^x\sec x$,

For $y = \dfrac{e^{2x}\sin x}{x}$, the numerator is itself a product. With $u = e^{2x}\sin x$ and $v = x$, the product rule gives $u' = 2e^{2x}\sin x + e^{2x}\cos x$, so the quotient rule gives

For $y = \ln(\sec x + \tan x)$, the chain rule for a logarithm gives

The last example shows a derivative that simplifies a long way once the common factor is spotted, and the same result is used in reverse when integrating $\sec x$.