Alongside integration, differentiation is the one of two main branches of calculus. It is used when finding the gradient of a curve as opposed to a straight line. Recall that the gradient of a straight line can be found by looking at its equation (the m in y=mx+c). Or it can be found by calculating rise over run. For a curve, however, the gradient is always changing. Thus the gradient of the curve at a given point is defined to be the gradient of the tangent at that point (see more on tangents). The function that tells us the gradient of the tangent/curve at the given point is known as the derivative. Since the gradient is dependent on x, the derivative should also be a function of x. For $y=f(x)$ the derivative is often denoted $\frac{dy}{dx}$ or $f$. Note that $\frac{dy}{dx}$ is the notation used for rise over run on a curve. It denotes an infinitely small change in x over an infinity small change in y. It represents the rate of change of y with respect to x. Once an expression for the derivative is obtained, it can be used to calculate the gradient at any point. See Polynomial Examples below.

## Differentiation from First Principles The formal technique for finding the gradient of a tangent is known as Differentiation from First Principles. 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. Evidently, the closer these points are together, the better the approximation. Suppose we want to find the gradient of the curve of $y=f(x)$ at a general point $(x,f(x))$. Choose another point close by $(x+h,f(x+h))$ where $h$ is small. Hence, 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: $f$

Note that this is the formal approach for finding the derivative. Shortcuts are, however, usually adopted. See Polynomial Differentiation below. Click here for a detailed explanation of Differentiation from First Principles on YouTube.

### 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 $f(x)=2x^3$ from first principles. $f$ $=\lim_{\delta x\rightarrow 0}(6x^2+6x\delta x+2\delta x^2)=6x^2$

Note that $\delta x$ is often used in place of h and denotes infinitesimal (infinitely small) change in x.

## Polynomial Differentiation

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 $\frac{dy}{dx}=3x^2$. It follows that the point (2,8) on the cubic graph has a gradient of 12. This is found by putting x=2 into the derivative.

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, $y=4x+3$ for example, is 4 – constants disappear. You may also be expected to simplify first (see Example 3).

### Example 1

Find $\frac{dy}{dx}$ if $y=4x^{2}-5x^{-1}+7x-2$.

Solution:

Using the above, $\frac{dy}{dx}=8x+5x^{-2}+7$

### Example 2

If $f(x)=8x^{-\frac{1}{2}}+3x^{\frac{1}{2}}$, find the gradient of the curve $y=f(x)$ at the point with x-coordinate 4.

Solution: Firstly, differentiating gives $f$. It follows that $f$ See Indices. Hence, the gradient of the curve at x=4 is 0.25.

### Example 3

Differentiate $h(x)=\frac{(2x^2-5)(x-2)}{\sqrt{x}}$.

Solution: Firstly, simplifying gives $h(x)=\frac{2x^3-5x-4x^2+10}{x^{\frac{1}{2}}}=2x^{\frac{5}{2}}-5x^{\frac{1}{2}}-4x^{\frac{3}{2}}+10x^{-\frac{1}{2}}$

Hence, $\frac{dh}{dx}=5x^{\frac{3}{2}}-\frac{5}{2}x^{-\frac{1}{2}}-6x^{\frac{1}{2}}-5x^{-\frac{3}{2}}$.

Click here to find Questions by Topic and scroll down to all past DIFFERENTIATION questions to practice some more. Now you’re ready to test your Pure Maths knowledge. So visit our Practice Papers page and take StudyWell’s own Pure Maths tests.