Answer :
U can with practical method or with definition.
[tex]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) [/tex]
[tex]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) [/tex]
What you need to know:
[tex]y=k{ x }^{ n }\\ \\ \ln { y } =\ln { \left( k{ x }^{ n } \right) } [/tex]
[tex]\\ \\ \ln { y } =\ln { k } +\ln { \left( { x }^{ n } \right) } \\ \\ \ln { y } =\ln { k } +n\ln { x }[/tex]
[tex]\\ \\ \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{ x }^{ -1 }\cdot k{ x }^{ n }\\ \\ \frac { dy }{ dx } =kn{ x }^{ -1+n }[/tex]
[tex]\\ \\ \frac { dy }{ dx } =kn{ x }^{ n-1 }[/tex]
If this is the case, when:
[tex]y=5{ x }^{ 2 }-2x+1[/tex]
dy/dx is...
[tex]\\ \\ \frac { dy }{ dx } =5\cdot 2{ x }^{ 2-1 }-2\cdot 1{ x }^{ 1-1 }\\ \\ \frac { dy }{ dx } =10x-2\cdot { x }^{ 0 }[/tex]
[tex]\\ \\ \frac { dy }{ dx } =10x-2[/tex]
[tex]y=k{ x }^{ n }\\ \\ \ln { y } =\ln { \left( k{ x }^{ n } \right) } [/tex]
[tex]\\ \\ \ln { y } =\ln { k } +\ln { \left( { x }^{ n } \right) } \\ \\ \ln { y } =\ln { k } +n\ln { x }[/tex]
[tex]\\ \\ \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{ x }^{ -1 }\cdot k{ x }^{ n }\\ \\ \frac { dy }{ dx } =kn{ x }^{ -1+n }[/tex]
[tex]\\ \\ \frac { dy }{ dx } =kn{ x }^{ n-1 }[/tex]
If this is the case, when:
[tex]y=5{ x }^{ 2 }-2x+1[/tex]
dy/dx is...
[tex]\\ \\ \frac { dy }{ dx } =5\cdot 2{ x }^{ 2-1 }-2\cdot 1{ x }^{ 1-1 }\\ \\ \frac { dy }{ dx } =10x-2\cdot { x }^{ 0 }[/tex]
[tex]\\ \\ \frac { dy }{ dx } =10x-2[/tex]
