Question

Use the definition of the derivative at a point to find the slope of the tangent line of f(x)=3x^2+2x+1 at x=1

Answer 1

By definition,

`f'(x) = \displaystyle \lim_(h\to0)\frac{f(x+h)-f(x)}h`

Let x = 1. Then

`f'(1) = \displaystyle \lim_(h\to0)\frac{f(1+h)-f(1)}h \\\\ f'(1) = \lim_(h\to0)\frac{(3(1+h)^2+2(1+h)+1)-6}h \\\\ f'(1) = \lim_(h\to0)\frac{3+6h+3h^2+2+h+1-6}h \\\\ f'(1) = \lim_(h\to0)\frac{7h+3h^2}h \\\\ f'(1) = \lim_(h\to0)(7+3h) = \boxed{7}`