Question

How to find the derivative 5x²-2x+1

Answer 1

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...

$\\ \\ \frac { dy }{ dx } =5\cdot 2{ x }^{ 2-1 }-2\cdot 1{ x }^{ 1-1 }\\ \\ \frac { dy }{ dx } =10x-2\cdot { x }^{ 0 }$

$\\ \\ \frac { dy }{ dx } =10x-2$

Answer 2

U can with practical method or with definition.

$f(x)=5x^2-2x+1\\ \\f'(x)=2*5x^{2-1}-1*2x^{1-1}+0\\ \\f'(x)=10x-2\\ \\f'(x)=2(5x-1)$