\(f'(x) = 3 ⋅ 2 ⋅ x^{1} \text{.}\)

\(f'(x) = 6 x \text{.}\)

\(f'(a) = -3 ⋅ 8 ⋅ a^{7} - 1 ⋅ 7 ⋅ a^{6} + 8 ⋅ 2 ⋅ a^{1} \text{.}\)

\(f'(a) = -24 a^{7} - 7 a^{6} + 16 a \text{.}\)

\(f'(p) = 1\frac{3}{4} ⋅ 7 ⋅ p^{6} + 1\frac{2}{3} ⋅ 6 ⋅ p^{5} + \frac{2}{9} ⋅ 4 ⋅ p^{3} + 1\frac{1}{3} ⋅ 2 ⋅ p^{1} \text{.}\)

\(f'(p) = 12\frac{1}{4} p^{6} + 10 p^{5} + \frac{8}{9} p^{3} + 2\frac{2}{3} p \text{.}\)

(Haakjes wegwerken)
\(f(a) = (3 a^{2} + 1) (a - 7) = 3 a^{3} - 21 a^{2} + a - 7\)

(Differentiëren)
\(f'(a) = 9 a^{2} - 42 a + 1 \text{.}\)

(Haakjes wegwerken)
\(f(x) = (2 x^{4} + 5)^{2} = 4 x^{8} + 20 x^{4} + 25\)

(Differentiëren)
\(f'(x) = 32 x^{7} + 80 x^{3} \text{.}\)

(Herleiden)
\(f(x) = -{2 \over 5 x^{7}} = -\frac{2}{5} x^{-7}\)

(Differentiëren)
\(f'(x) = -\frac{2}{5} ⋅ -7 ⋅ x^{-8} = \frac{14}{5} ⋅ x^{-8}\)

(Herleiden)
\(f'(x) = \frac{14}{5} ⋅ {1 \over x^{8}} = {14 \over 5 x^{8}}\)

(Uitdelen)
\(f(p) = {p^{6} \over 3 p^{4}} - {2 p^{3} \over 3 p^{4}} = \frac{1}{3} p^{2} - \frac{2}{3} p^{-1}\)

(Differentiëren)
\(f'(p) = \frac{1}{3} ⋅ 2 ⋅ p - \frac{2}{3} ⋅ -1 ⋅ p^{-2}\)

(Herleiden)
\(f'(p) = \frac{2}{3} p + {2 \over 3 p^{2}}\)

(Herleiden)
\(f(a) = {5 \over 8 \sqrt{a}} + 6 \sqrt{a} = \frac{5}{8} a^{-\frac{1}{2}} + 6 a^{\frac{1}{2}}\)

(Differentiëren)
\(f'(a) = \frac{5}{8} ⋅ -\frac{1}{2} ⋅ a^{-1\frac{1}{2}} + 6 ⋅ \frac{1}{2} ⋅ a^{-\frac{1}{2}}\)

(Herleiden)
\(f'(a) = -{5 \over 16 a \sqrt{a}} + {3 \over \sqrt{a}}\)

(Kettingregel)
\(f'(a) = 5 ⋅ 9 ⋅ (a + 4)^{8} ⋅ 1\)

(Herleiden)
\(f'(a) = 45 (a + 4)^{8} \text{.}\)

(Herleiden)
\(f(x) = {5 \over (2 x + 1)^{4}} = 5 ⋅ (2 x + 1)^{-4}\)

(Kettingregel)
\(f'(x) = 5 ⋅ -4 ⋅ (2 x + 1)^{-5} ⋅ 2\)

(Herleiden)
\(f'(x) = -40 ⋅ (2 x + 1)^{-5} = -{40 \over (2 x + 1)^{5}}\)

(Herleiden)
\(f(a) = \frac{7}{8} \sqrt{4 a + 2} = \frac{7}{8} ⋅ (4 a + 2)^{\frac{1}{2}} \text{.}\)

(Kettingregel)
\(f'(a) = \frac{7}{8} ⋅ \frac{1}{2} ⋅ (4 a + 2)^{-\frac{1}{2}} ⋅ 4\)

(Herleiden)
\(f'(a) = \frac{7}{4} ⋅ (4 a + 2)^{-\frac{1}{2}} = {7 \over 4 \sqrt{4 a + 2}}\)

(Herleiden)
\(f(p) = {6 \over 7 \sqrt{3 p - 1}} = \frac{6}{7} ⋅ (3 p - 1)^{-\frac{1}{2}}\)

(Kettingregel)
\(f'(p) = \frac{6}{7} ⋅ -\frac{1}{2} ⋅ (3 p - 1)^{-1\frac{1}{2}} ⋅ 3\)

(Herleiden)
\(f'(p) = -\frac{9}{7} ⋅ (3 p - 1)^{-1\frac{1}{2}} = -{9 \over 7 (3 p - 1) \sqrt{3 p - 1}}\)