Ответ:
Объяснение:
а) f(x) = x¹² ; f'(x) = 12x¹¹
б) f(x) = -√5 х⁶ ; f'(x) = - 6√5x⁵
в) f(x)= 2x⁸ - 11x⁵ - 4x² + 26 ; f'(x) = 2*8х⁷ -11*5х⁴ + 4*2х = 16х⁷ - 55х⁴ +8х
г) f(x) = (x³-2) ∙ (x²+3); f’(x) = [(uv)’= uv’+u’v; u = x² +3; v=x³-2] =
(x³-2)(x²+3)’ + (x³-2)’(x²+3) = (x³-2)2x + (x²+3)3x² = 5x⁴ +9x² - 4x
д) f(x) = -x/(x² + 9)
![\displaystyle f'(x) = \left[\begin{array}{ccc}(\frac{u}{v})'=\frac{vu'-uv'}{v^{2} } \\u=x\\v=x^{2} +9\end{array}\right] =- \frac{(x^{2} +9)(x)'-x(x^{2}+9)' }{(x^{2}+9 )^2} =](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28%5Cfrac%7Bu%7D%7Bv%7D%29%27%3D%5Cfrac%7Bvu%27-uv%27%7D%7Bv%5E%7B2%7D%20%7D%20%20%5C%5Cu%3Dx%5C%5Cv%3Dx%5E%7B2%7D%20%2B9%5Cend%7Barray%7D%5Cright%5D%20%3D-%20%5Cfrac%7B%28x%5E%7B2%7D%20%2B9%29%28x%29%27-x%28x%5E%7B2%7D%2B9%29%27%20%7D%7B%28x%5E%7B2%7D%2B9%20%29%5E2%7D%20%3D "\displaystyle f'(x) = \left[\begin{array}{ccc}(\frac{u}{v})'=\frac{vu'-uv'}{v^{2} } \\u=x\\v=x^{2} +9\end{array}\right] =- \frac{(x^{2} +9)(x)'-x(x^{2}+9)' }{(x^{2}+9 )^2} =")
