Question

What is the derivative of 2x squared

Answer 1

Firstly,


`y={ kx }^( n )\\ \\ \ln { y } =\ln { \left( k{ x }^( n ) \right) } \\ \\ \ln { y } =\ln { k } +\ln { \left( { x }^( n ) \right) } \\ \\ \ln { y } =\ln { k } +n\cdot \ln { x } \\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { n }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { n }{ x } \cdot y`


`\\ \\ \frac { dy }{ dx } =n\cdot { x }^( -1 )\cdot k\cdot { x }^( n )\\ \\ \therefore \quad \frac { dy }{ dx } =kn{ x }^( n-1 )`

*Keep this proof in your reference book, because you may need it in the future.

---------------------

WHAT YOU NEED TO KNOW:


`If:\quad y={ kx }^( n )\\ \\ \therefore \quad \frac { dy }{ dx } =kn{ x }^( n-1 )`

This means that:


`As:\quad y={ 2x }^( 2 )\\ \\ k=2\quad and\quad n=2,\\ \\ \therefore \quad \frac { dy }{ dx } =4{ x }`

Answer:

4x