Question

(6х7-2х5+10)'
$(4 \sqrt{ \times )} .$
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Answer 1

$3)\; \; (4x^7-2x^5+19)'=6\cdot (x^7)'-2\cdot (x^5)'+10'=\\\\=6\cdot 7x^6-2\cdot 5x^4+0=42x^6-10x^4\\\\\\4)\; \; \Big (\frac{5}{x^5}\Big )'=-\frac{5\cdot (x^5)'}{(x^5)^2}=-\frac{25x^4}{x^{10}}=-\frac{25}{x^6}\\\\ili\; \; \; \Big (\frac{5}{x^5}\Big )'=(5x^{-5})'=5\cdot (-5)x^{-6}=-\frac{25}{x^6}$