\[4x^3\]
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
\[\begin{equation} \begin{aligned} \frac{dy}{dx} &=4 \times 2x^{2-1}-6 \times x^{1-1}-4 \times 0 x^{0-1} \\ & =8x^1-6 \times x^0 \\ &= 8x-6 \times 1 \\ &= 8x-6\end{aligned} \end{equation}\]