(A) f(x) = 7 is constant, so f(x + h) = 7, too, which makes f(x + h) - f(x) = 0. So f'(x) = 0.
(B) f(x) = 5x + 1 ==> f(x + h) = 5 (x + h) + 1 = 5x + 5h + 1
==> f(x + h) - f(x) = 5h
Then
![\displaystyle f'(x) = \lim_(h\to0)\frac{5h}h = \lim_(h\to0)5 = 5](https://img.qammunity.org/2022/formulas/mathematics/college/cztg1p4q9imrgtmifovq9v3i3v60pdt6g3.png)
(C) f(x) = x ² + 3 ==> f(x + h) = (x + h)² + 3 = x ² + 2xh + h ² + 3
==> f(x + h) - f(x) = 2xh + h ²
![\implies\displaystyle f'(x) = \lim_(h\to0)\frac{2xh+h^2}h = \lim_(h\to0)(2x+h) = 2x](https://img.qammunity.org/2022/formulas/mathematics/college/xdeuu6ncu442dah5cgvd9u3c3y0mnu9ft2.png)
(D) f(x) = x ² + 4x - 1 ==> f(x + h) = (x + h)² + 4 (x + h) - 1 = x ² + 2xh + h ² + 4x + 4h - 1
==> f(x + h) - f(x) = 2xh + h ² + 4h
![\implies \displaystyle f'(x) = \lim_(h\to0)\frac{2xh+h^2+4h}h = \lim_(h\to0)(2x+h+4) = 2x+4](https://img.qammunity.org/2022/formulas/mathematics/college/k7q71vthy9ij7zx0zob7w33lvyzbeshj3l.png)