Firstly,
[tex]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[/tex]
[tex]\\ \\ \frac { dy }{ dx } =n\cdot { x }^{ -1 }\cdot k\cdot { x }^{ n }\\ \\ \therefore \quad \frac { dy }{ dx } =kn{ x }^{ n-1 }[/tex]
*Keep this proof in your reference book, because you may need it in the future.
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WHAT YOU NEED TO KNOW:
[tex]If:\quad y={ kx }^{ n }\\ \\ \therefore \quad \frac { dy }{ dx } =kn{ x }^{ n-1 }[/tex]
This means that:
[tex]As:\quad y={ 2x }^{ 2 }\\ \\ k=2\quad and\quad n=2,\\ \\ \therefore \quad \frac { dy }{ dx } =4{ x }[/tex]
Answer:
4x
