Question

найти производную/знайти похідну ​

Answer 1

1)

$f(x) = {x}^{5} + 3 {x}^{2} + 7 \\ f'(x) = 5 {x}^{5 - 1} + 3 \times 2 {x}^{2 - 1 } = 5 {x}^{4} + 6x$

2)

$f(x) = \frac{ {x}^{4} }{4} - \frac{ {x}^{3} }{6} - 7x + 3 \\ f'(x) = \frac{4}{4} {x}^{4 - 1} - \frac{3}{6} x {}^{3 - 1} - 7x {}^{1 - 1} = \\ {x}^{3} - 0.5 {x}^{2} - 7$

3)

$f(x) = {x}^{4} (2x {}^{2} - 3) = 2 {x}^{6} - 3 {x}^{4} \\ f'(x) = 2 \times 6 {x}^{6 - 1} - 3 \times 4 {x}^{4 - 1} = \\ 12 {x}^{5} - 12 {x}^{3} = 12 {x}^{3} (x {}^{2} - 1) = \\ 12 {x}^{3} (x - 1)(x + 1)$

4)

$f(x) = \frac{2x + 5}{ {x}^{2} + 1 } \\ f'(x) = \frac{(2x + 5)'( {x}^{2} + 1) \times ( {x}^{2} + 1)'(2x + 5)}{ ({x}^{2} + 1) {}^{2} } = \\ \frac{2( {x}^{2} + 1) - 2x(2x + 5) }{( {x}^{2} + 1) {}^{2} } = \\ \frac{2 {x}^{2} + 2 - 4 {x}^{2} - 10x }{( {x}^{2} + 1) {}^{2} } = \frac{ - 2 {x}^{2} + 10x + 2 }{(x {}^{2} + 1) {}^{2} }$