Question

f(x)=2x^2+10x, a=3. Find the equation of the tangent line, (I am guessing based on this equation: y-f(a)=f'(a)(x-a) where a=3) Therefore: y-f(3)=f'(3)(x-3). Then the first derivative would be f'(x)=f(3+h)-f(3)/h

Answer 1

there's an easier way to calculate derivative
f'(x+y+z)=f'(x)+f'(y)+f'(z)
so f'(2x^2+10x)=4x+10
f'(3)=4(3)+10=22

Answer 2

I know there is always an easier way, but this is the way the hw told me to do it.

Answer 3

\[\lim_{h \rightarrow 0} [2 \times (x+h)^2+ 10 \times (x+h)]-(2x^2+10x)/h\]
\[=\lim_{h \rightarrow 0} (2x^2+4xh+2h^2+10x+10h-2x^2-10x)/h \]
\[\lim_{h \rightarrow 0} (4xh+2h^2+10h)/h=\lim_{h \rightarrow 0} h(4x+2h+10)/h=\lim_{h \rightarrow 0} 4x+2h+10 = 4x+10\]