Use First Principles to find the value of the derivative when x = 1

Question

f(x) = 2x² + 3x + 1

i need some help finding the derivative please!

Answer 1

The limit definition of a derivative (using first principles) is such that

f'(x) = lim_h->0 (f(x + h) - f(x))/h

Since f(x) = 2x² + 3x + 1, f(x + h) = 2(x + h)² + 3(x + h) + 1 = 2x² + 4xh + 2h² + 3x + 3h + 1.

Then, f(x + h) - f(x) = 2x² + 4xh + 2h² + 3x + 3h + 1 - 2x² - 3x - 1 = 2h² + 4xh + 3h

We can divide the entire expression by h so that

(f(x + h) - f(x))/h = (2h² + 4xh + 3h)/h = 2h + 4x + 3

And finally, we take the limit as h approaches 0 such that

f'(x) = lim_h->0 (f(x + h) - f(x))/h = lim_h->0 2h + 4x + 3 = 4x + 3

Thus, f'(x) = 4x + 3. Evaluating the derivative at x = 1 yields

f'(1) = 4(1) + 3 = 7