SOLUTION
The given function is
[tex]f(x)=7x-3x^2[/tex]
Using the limit definition
[tex]f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]
Substitute x+h for x
This gives
[tex]f^{\prime}(x)=\lim _{h\to0}\frac{7(x+h)+3(x+h)^2-(7x-3x^2)}{h}[/tex]
Simplify the limit
[tex]\begin{gathered} f^{\prime}(x)=\lim _{h\to0}\frac{7x+7h-3(x^2+2hx+h^2)^{}-7x+3x^2}{h} \\ \end{gathered}[/tex]
This further gives
[tex]f^{\prime}(x)=\lim _{h\to0}\frac{7x+7h-3x^2-6hx-3h^2-7x+3x^2}{h}[/tex]
Simplify further
[tex]f^{\prime}(x)=\lim _{h\to0}\frac{7h-6hx-3h^2}{h}[/tex]
Simplify the fraction
[tex]\begin{gathered} f^{\prime}(x)=\lim _{h\to0}(\frac{7h}{h}-\frac{6hx}{h}-\frac{3h^2}{h}) \\ f^{\prime}(x)=\lim _{h\to0}(7-6x-3h) \end{gathered}[/tex]
Find the limit
[tex]\begin{gathered} f^{\prime}(x)=(7-6x-3(0)) \\ f^{\prime}(x)=7-6x \end{gathered}[/tex]
Therefore, the solution is
[tex]f^{\prime}(x)=7-6x[/tex]
