What you need to know:
\(y=k{ x }^{ n }\\ \\ \ln { y } =\ln { \left( k{ x }^{ n } \right) }\)
\(\\ \\ \ln { y } =\ln { k } +\ln { \left( { x }^{ n } \right) } \\ \\ \ln { y } =\ln { k } +n\ln { x }\)
\(\\ \\ \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { n }{ x } \\ \\ y\cdot \frac { 1 }{ y } \cdot \frac { dy }{ dx } =\frac { n }{ x } \cdot y\)
\(\\ \\ \frac { dy }{ dx } =n{ x }^{ -1 }\cdot k{ x }^{ n }\\ \\ \frac { dy }{ dx } =kn{ x }^{ -1+n }\)
\(\\ \\ \frac { dy }{ dx } =kn{ x }^{ n-1 }\)
If this is the case, when:
\(y=5{ x }^{ 2 }-2x+1\)
dy/dx is...
\(\\ \\ \frac { dy }{ dx } =5\cdot 2{ x }^{ 2-1 }-2\cdot 1{ x }^{ 1-1 }\\ \\ \frac { dy }{ dx } =10x-2\cdot { x }^{ 0 }\)
\(\\ \\ \frac { dy }{ dx } =10x-2\)
