Question

математика
помогите пожалуйста ​

Answer 1

Ответ:

$\dfrac{4x^{2}-6x}{x^{4}};$

$\dfrac{-8x^{7}+48x^{5}+3x^{2}-3}{(1+4x^{5})^{2}};$

Пошаговое объяснение:

$\bigg (\dfrac{u}{v} \bigg )'=\dfrac{u'v-uv'}{v^{2}};$

$\bigg (\dfrac{3-4x}{x^{2}} \bigg )'=\dfrac{(3-4x)' \cdot x^{2}-(3-4x) \cdot (x^{2})'}{(x^{2})^{2}}=\dfrac{-4 \cdot x^{2}-(3-4x)\cdot 2x}{x^{2 \cdot 2}}=$

$=\dfrac{-4x^{2}-6x+8x^{2}}{x^{4}}=\dfrac{4x^{2}-6x}{x^{4}};$

$\bigg (\dfrac{x^{3}-3x}{1+4x^{5}} \bigg )'=\dfrac{(x^{3}-3x)' \cdot (1+4x^{5})-(x^{3}-3x) \cdot (1+4x^{5})'}{(1+4x^{5})^{2}}=$

$=\dfrac{(3x^{2}-3) \cdot (1+4x^{5})-(x^{3}-3x) \cdot 20x^{4}}{(1+4x^{5})^{2}}=\dfrac{3x^{2}+12x^{7}-3-12x^{5}-20x^{7}+60x^{5}}{(1+4x^{5})^{2}}=$

$=\dfrac{-8x^{7}+48x^{5}+3x^{2}-3}{(1+4x^{5})^{2}};$