A-level Mathematics/CIE/Pure Mathematics 2/Differentiation

Differentiating Logarithmic and Exponential functions

The function $y=e^x$ is its own derivative: $\frac{d}{dx}{e}^x=e^x$. The constant $e$ is defined such that this is true.

An exponential function with a different base can be converted into a function of the form $e^{kx}$ using logarithms, e.g. $2^x=e^{\ln 2x}$. The derivative of such an expression can be found using the chain rule: $\frac{d}{dx}{2}^x=\frac{d}{dx}{e}^{\ln 2x}=\ln 2e^{\ln 2x}=2^x\ln 2$.

$\frac{d}{dx}{e}^{f(x)}=f^{\prime }(x)e^{f(x)}$

The derivative of a logarithm is $\frac{d}{dx}\ln x=\frac{1}{x}$. Applying the chain rule to this produces the result:

$\frac{d}{dx}\ln (f(x))=\frac{f^{\prime }(x)}{f(x)}$

It is important to know how these rules interact with other expressions.

e.g. $\frac{d}{dx}{e}^{x^2+\ln \frac{1}{x}}=e^{x^2+\ln \frac{1}{x}}(\frac{d}{dx}{x}^2+\ln \frac{1}{x})=e^{x^2+\ln \frac{1}{x}}(2x+\frac{-x^{-2}}{1/x})=e^{x^2+\ln \frac{1}{x}}(2x-x^{-1})$

Differentiating Trigonometric Functions

The trigonometric functions have the following derivatives:

• $\frac{d}{dx}\sin x=\cos x$
• $\frac{d}{dx}\cos x=-\sin x$
• $\frac{d}{dx}\tan x={\sec }^{2}x$
• $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

The Product Rule

The product rule states that:

$\frac{d}{dx}f(x)g(x)=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$

e.g. $\frac{d}{dx}x{e}^x=x\frac{d}{dx}{e}^x+e^x\frac{d}{dx}x=x{e}^x+e^x=e^x(x+1)$

The Quotient Rule

The quotient rule is a special case of the product rule when one of the terms in the product is a reciprocal.

e.g. Evaluate $\frac{d}{dx}\frac{2x+3}{x+4}$

$\begin{array}{cc}\frac{d}{dx}\frac{2x+3}{x+4}& =\frac{d}{dx}(2x+3)\frac{1}{x+4}\\ & =(2x+3)\frac{d}{dx}\left(\frac{1}{x+4}\right)+\frac{1}{x+4}\frac{d}{dx}(2x+3)\\ & =-(2x+3)(x+4{)}^{-2}+\frac{1}{x+4}(2)\\ & =\frac{-(2x+3)}{(x+4{)}^2}+\frac{2}{x+4}\\ & =\frac{-(2x+3)}{(x+4{)}^2}+\frac{2x+8}{(x+4{)}^2}\\ & =\frac{2x+8-2x-3}{(x+4{)}^2}\\ & =\frac{5}{(x+4{)}^2}\end{array}$

In general:

$\begin{array}{cc}\frac{d}{dx}\frac{u}{v}& =u\frac{d}{dx}\frac{1}{v}+\frac{1}{v}\frac{du}{dx}\\ & =u(-v{)}^{-2}\frac{dv}{dx}+\frac{v}{v^2}\frac{du}{dx}\\ & =\frac{-u\frac{dv}{dx}+v\frac{du}{dx}}{v^2}\\ & =\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\end{array}$

Implicit Differentiation

Implicit differentiation is where we differentiate a function which is not defined explicitly, with y as the subject. To do this, it is sensible to use the chain rule.

e.g. Find an expression for $\frac{dy}{dx}$ when $x^2+y^2=1$.

$\begin{array}{cc}{x}^2+y^2& =1\\ \frac{d}{dx}{x}^2+y^2& =\frac{d}{dx}1\\ 2x+\frac{d{y}^2}{dx}& =0\\ 2x+\frac{d{y}^2}{dy}\frac{dy}{dx}& =0\text{Use the chain rule}\\ 2x+2y\frac{dy}{dx}& =0\\ 2y\frac{dy}{dx}& =-2x\\ \frac{dy}{dx}& =-\frac{2x}{2y}=-\frac{x}{y}\end{array}$

Sometimes, we need to use the product rule too.

e.g. Find an expression for $\frac{dy}{dx}$ when $x^2+6xy+y^2=1$.

$\begin{array}{cc}{x}^2+6xy+y^2& =1\\ \frac{d}{dx}{x}^2+6xy+y^2& =\frac{d}{dx}1\\ 2x+\frac{d}{dx}6xy+\frac{d}{dx}{y}^2& =0\\ 2x+6x\frac{dy}{dx}+y\frac{d}{dx}6x+\frac{d{y}^2}{dy}\frac{dy}{dx}& =0\\ 2x+6x\frac{dy}{dx}+6y+2y\frac{dy}{dx}& =0\\ 2x+6y+(6x+2y)\frac{dy}{dx}& =0\\ (6x+2y)\frac{dy}{dx}& =-2x-6y\\ \frac{dy}{dx}& =\frac{-2x-6y}{6x+2y}\\ \frac{dy}{dx}& =\frac{-x-3y}{3x+y}\end{array}$

Parametric Differentiation

A parametric function is where instead of $y$ being defined by $x$, $y$ and $x$ are both linked to a third parameter, $t$. e.g. $x=3t,y=t^2$

To find $\frac{dy}{dx}$ when $y$ and $x$ are defined parametrically, we need to use the chain rule:

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}$

So for the example $x=3t,y=t^2$, $\frac{dx}{dt}=3$ and $\frac{dy}{dt}=2t$, thus $\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}=2t÷3=\frac{2t}{3}$