using limit definition
[tex]\begin{gathered} F(x)=10-2x-x^2 \\ F^1(x)\text{=}\lim _{\Delta x-0}\frac{\text{f(x+}\Delta x)-f(x)}{\Delta x} \end{gathered}[/tex][tex]f^1(x)=\lim _{\Delta x-0}\frac{10-2(x+\Delta x)-(x+\Delta x)^2-(10-2x-x^2)}{\Delta x}[/tex]
Next is to expand the numerator of the fraction
[tex]f^1(x)=\lim _{\Delta x-0}\frac{10-2x-2\Delta x-x^2-2x\Delta x-(\Delta x)^2-10+2x+x^2}{\Delta x}[/tex]
You now divide through by Δx and simplify the expression
[tex]f^1(x)=\lim _{\Delta x-o}\frac{10-10-2x+2x-x^2+x^2}{\Delta x}+\frac{-2\Delta x-2x\Delta x-(\Delta x)^2}{\Delta x}[/tex]
The first fraction becomes 0 and you divide through by Δx in the second fraction
[tex]\begin{gathered} f^!(x)=\lim _{\Delta x-0}(-2-2x-\Delta x) \\ \text{Substitute }\Delta x\text{ =0} \\ f^1(x)=-2-2x \\ =-2(1+x) \end{gathered}[/tex]
Hence the derivative of f(x) is -2(1+x)
