Differentiation

To differentiate: Times by the power and take one off the power.
$4x^3$
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{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}