Question

Найти производную функции f(x)=(3-x3)5+под корнем 2х-7

Answer 1

Предположу, что
$f(x) = {(3 - {x}^{3} })^{5} + \sqrt{2x - 7}$
Тогда,
f'(x) =
$=5 {(3 - {x}^{3} })^{4} ( - 3 {x}^{2} ) + \\ + \frac{1}{2 \sqrt{2x - 7} } \times 2 = \\ = - 15 {x}^{2} {(3 - {x}^{3} })^{4} + \\ +\frac{1}{ \sqrt{2x - 7} }$

Answer 2

$\mathtt{f'(x)=[(3-x^3)^5]'+[\sqrt{2x-7}]'=5(3-x^3)'(3-x^3)^4+\frac{(2x-7)'}{2\sqrt{2x-7}}=}\\\mathtt{5*(-3x^2)(3-x^3)^4+\frac{2}{2\sqrt{2x-7}}=-15x^2(3-x^3)^4+(2x-7)^{-\frac{1}{2}}}$