Question

допоможіть знайти похідну

Answer 1

Ответ:

10x⁴tgx – (x+(arccos x)√(1 – x²))/(x²√(1 – x²)) + (3 + 2x⁵)/(cos²x)

$10x^{4}tgx-\dfrac{x+(\arccos x) \sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}+\dfrac{3+2x^{5}}{\cos^{2}x}$

Пошаговое объяснение:

y = (arccos x)/x + (3 + 2x⁵)tg x – 1 ;

$y=\dfrac{\arccos x}{x}+(3+2x^{5})tgx-1;$

$\dfrac{dy}{dx}=\bigg (\dfrac{\arccos x}{x}+(3+2x^{5})tgx-1 \bigg)'=\bigg (\dfrac{\arccos x}{x} \bigg )'+((3+2x^{5})tgx)'-1'=$

$=\dfrac{(\arccos x)' \cdot x-\arccos x \cdot x'}{x^{2}}+(3+2x^{5})' \cdot tgx+(3+2x^{5}) \cdot (tgx)'-0=$

$=\dfrac{-\dfrac{1}{\sqrt{1-x^{2}}} \cdot x-\arccos x \cdot 1}{x^{2}}+(3'+(2x^{5})') \cdot tgx+(3+2x^{5}) \cdot \dfrac{1}{\cos^{2}x}=$

$=\dfrac{-\dfrac{x}{\sqrt{1-x^{2}}}-\arccos x}{x^{2}}+(0+2 \cdot (x^{5})') \cdot tgx+\dfrac{3+2x^{5}}{\cos^{2}x}=-\dfrac{\dfrac{x}{\sqrt{1-x^{2}}}+\arccos x}{x^{2}}+$

$+2 \cdot 5 \cdot x^{5-1} \cdot tgx+\dfrac{3+2x^{5}}{\cos^{2}x}=10x^{4}tgx-\dfrac{x+(\arccos x) \sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}+\dfrac{3+2x^{5}}{\cos^{2}x};$