## Differentiation from First Principles

Recall that when finding the gradient of a straight line, one can choose two points on the line and calculate the rise over run. When finding the gradient of a point on a curve, one must find the gradient of the tangent at that point. This can be done by calculating rise over run for two points on the tangent but these will not be points that lie on the curve. By taking two points on the curve that lie very closely together, the straight line between them will have approximately the same gradient as the tangent there.

Suppose we want to find the gradient of the curve of $y=f(x)$ at a general point $x$. Take another point close by $x+h$ where $h$ is small. The two points have corresponding $y$ coordinates $f(x)$ and $f(x+h)$ and so rise over run is given by $\frac{f(x+h)-f(x)}{x+h-x}$. By taking $h$ to be as small as possible, i.e. the limit as $h$ tends to 0, then rise over run will be the same as the gradient of the tangent. This is known as DIFFERENTIATION FROM FIRST PRINCIPLES:

$f$

### Example 1

Differentiate $y=x^2$ from first principles.

$\frac{dy}{dx}=\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}=\lim_{h\rightarrow 0}\frac{x^2+2xh+h^2-x^2}{h}=\lim_{h\rightarrow 0}(2x+h)=2x$

### Example 2

Differentiate $y=2x^3$ from first principles.

$\frac{dy}{dx}=\lim_{h\rightarrow 0}\frac{2(x+h)^3-2x^3}{h}=\lim_{h\rightarrow 0}\frac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}$
$=\lim_{h\rightarrow 0}(6x^2+6xh+2h^2)=6x^2$

## Differentiating Polynomials

It can be shown by differentiating from first principles that

$\frac{d}{dx}\left(x^n\right)=nx^{n-1}$

For example, if $y=x^3$ then $\frac{dy}{dx}=3x^2$ and it follows that the point (2,8) on the cubic graph has a gradient of 12.

The principle can be applied to linear combinations of powers of $x$, also known as polynomials. For example, the polynomial $f(x)=5x^2-2x^4$ has the derivative $f$. One can think of it informally as ‘multiplying down by the power, then taking one off of the power’.

Note that, in maths, differentiation is finding the derivative or gradient function. The gradient of the straight line, $y=4$ for example, is zero and so the derivative of a constant is 0.

Example 1:
$y=4x^{2}-5x^{-1}+7, \frac{dy}{dx}=8x+5x^{-2}$

Example 2:
$f(x)=8x^{-\frac{1}{2}}+3x^{\frac{1}{2}}, f$

## Differentiating e to the kx

The function $y=e^{kx}$, where k is a constant, can be found using the chain rule. It is useful to remember the following:

$y=e^{kx},\hspace{10pt}\frac{dy}{dx}=ke^{kx}$

For k=1, this says that at each point on the graph of y, the gradient matches that of the y coordinate:

$y=e^{x},\hspace{10pt}\frac{dy}{dx}=e^{x}$

Example 1Differentiate $y=2e^{4x}$.

$\frac{dy}{dx}=4\times 2e^{4x}=8e^{4x}$

Example 2Given that

$h(x)=\frac{1+20e^{5x}}{4e^{2x}}$.

Find h'(x).

h(x) can be written as

$h(x)=\frac{1}{4e^{2x}}+\frac{20e^{5x}}{4e^{2x}}=\frac{1}{4}e^{-2x}+5e^{3x}.$

We can differentiate more simply when h is written in this format:

$h$

Example 3 After time t seconds, the temperature, T degrees, of a heated metal ball that is dropped into a liquid is given by

$T(t)=350e^{-0.1t}+27$

Find the temperature of the ball at the instant it is dropped into the liquid. Find the rate at which the ball is cooling after 10 seconds.

Substituting t=0 into T gives the intial temperature as 377 degrees. The rate of cooling is given by the derivative of T:

$T$

At time t=10 seconds, the rate of cooling is

$T$

to 2 decimal places. This means that the ball is cooling by 12.88 degrees per second but this is only true at the instant 10 seconds after the ball is dropped into the liquid.

See more on Growth & Decay.