## Differentiation

\[4x^3\]

differentiated is \[4 \times 3 x^{3-1}=12x^2\]

We can differentiate a sum usinf the same rule for each term.

\[2x^5-4x^7\]

when differentiated is \[2 \times 5x^{5-1}-4 \times 7x^{7-1}=10x^4-8x^6\]

.This rule 'times by the power and take one off the power' works for

\[x\]

's and constants too.To differentiate

\[3x\]

write as \[3x^1\]

then apply the rule to give \[3 \times 1x^{1-1}=3x^0=3 \times 1=1\]

since \[x^0=1\]

.To differentiate

\[4\]

write as \[4x^0\]

then differentiatie using the above gives \[4 \times 0 x^{0-1}=0\]

since anything times 0 equals 0.Finally remember that when you differentiate

\[y\]

you get \[\frac{dy}{dx}\]

.Differentiate

\[y=4x^2-6x-4\]

.Write

\[y=4x^2-6x^1-4x^0\]

.We have

\[\begin{equation} \begin{aligned} \frac{dy}{dx} &=4 \times 2x^{2-1}-6 \times x^{1-1}-4 \times 0 x^{0-1} \\ & =8x^1-6 \times x^0 \\ &= 8x-6 \times 1 \\ &= 8x-6\end{aligned} \end{equation}\]