# A-level Mathematics/MEI/C3/Differentiation

< A-level Mathematics‎ | MEI‎ | C3

Differentiation in Core 3 (C3) are an extension of the work that you did in Core 1 and Core 2.

# Differentiation

## Standard Derivatives

For the C3 module, there are a few standard results for differentiation that need to be learnt. These are:

${\displaystyle {\frac {d}{dx}}\ln x={\frac {1}{x}}}$

${\displaystyle {\frac {d}{dx}}e^{kx}=ke^{kx}}$

${\displaystyle {\frac {d}{dx}}\sin kx=k\cos kx}$

${\displaystyle {\frac {d}{dx}}\cos kx=-k\sin kx}$

${\displaystyle {\frac {d}{dx}}\tan kx={\frac {k}{cos^{2}{kx}}}}$

## Chain Rule

${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}{\frac {du}{dx}}}$

The Chain Rule is used to differentiate when one function is applied to another function. A typical example of this is:

${\displaystyle y=\sin(x^{2})}$

One of the ways of remembering the chain rule is: Find the derivative outside, then multiply it by the derivative inside. In the example above, this becomes:

${\displaystyle {\frac {dy}{dx}}=2x\cos(x^{2})}$

## Product Rule

${\displaystyle {\frac {d}{dx}}uv=v{\frac {du}{dx}}+u{\frac {dv}{dx}}}$

The product rule is used when two functions are multiplied together.

## Quotient Rule

${\displaystyle {\frac {d}{dx}}{\frac {u}{v}}={\cfrac {v{\cfrac {du}{dx}}-u{\cfrac {dv}{dx}}}{v^{2}}}}$

The quotient rule is used when one function is divided by another. It is a specific case of the product rule. A typical example of this is:

## Implicit Differentiation

Implicit differentiation is used when a function is not a simple ${\displaystyle y=something}$ but contains a mixture of x and y parts. A typical example of this is to differentiate:

${\displaystyle y^{2}+2y=4x^{3}}$

When differentiating the y components of the expression you differentiate as normal, and then multiply by ${\displaystyle {\frac {dy}{dx}}}$. So differentiating both sides of the above expression it becomes:

${\displaystyle 2y{\frac {dy}{dx}}+2{\frac {dy}{dx}}=12x^{2}}$

Then by factorising the left hand side and cancelling, this becomes:

${\displaystyle {\frac {dy}{dx}}={\frac {6x^{2}}{y+1}}}$