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

Differentiating Logarithmic and Exponential functions
The function $$y = e^x$$ is its own derivative: $$\dfrac{d}{dx} e^x = e^x$$. 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. $$2^x = e^{\ln 2 x}$$. The derivative of such an expression can be found using the chain rule: $$\dfrac{d}{dx} 2^x = \dfrac{d}{dx} e^{\ln 2 x} = \ln 2\ e^{\ln 2 x} = 2^x \ln 2$$.

$$\dfrac{d}{dx} e^{f(x)} = f'(x)e^{f(x)}$$

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

$$\dfrac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$$

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

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

Differentiating Trigonometric Functions
The trigonometric functions have the following derivatives:


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

The Product Rule
The product rule states that:

$$\dfrac{d}{dx} f(x)g(x) = f(x)g'(x) + g(x)f'(x)$$

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

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 $$\dfrac{d}{dx} \frac{2x+3}{x+4} $$

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

In general:

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

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 $$\dfrac{dy}{dx}$$ when $$x^2 + y^2 = 1$$.

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

Sometimes, we need to use the product rule too.

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

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

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

$$\dfrac{dy}{dx} = \dfrac{dy}{dt}\times\dfrac{dt}{dx} = \dfrac{dy}{dt}\div\dfrac{dx}{dt}$$

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