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