Найти производную 1)y=ln(arcsin²5x) 2)y=x³·sin⁵3x+tg√x. Created by marinka31. algebra-ru

Найти производную 1)y=ln(arcsin²5x) 2)y=x³·sin⁵3x+tg√x...

marinka31 @marinka31

July 2022 1 8 Report

Найти производную 1)y=ln(arcsin²5x)
2)y=x³·sin⁵3x+tg√x

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nafanya2014

Verified answer

1)y`=(ln(arcsin²5x))`=

$=\frac{1}{arcsin^25x}\cdot(arcsin^25x)`= \frac{1}{arcsin^25x}\cdot 2arcsin5x\cdot(arcsin5x)`= \\ \\ =\frac{1}{arcsin^25x}\cdot 2arcsin5x\cdot \frac{1}{ \sqrt{1-(5x)^2} }\cdot(5x)` =\frac{10}{arcsin5x\cdot \sqrt{1-25x^2} }.$

2)y`=(x³·sin⁵3x)`+(tg√x)`=

$=3x^{2}\cdot sin^53x+x^3\cdot 5sin^43x\cdot(sin3x)`+ \frac{1}{cos^2 \sqrt{x} }\cdot( \sqrt{x} )`= \\ \\ =3x^{2}\cdot sin^53x+x^3\cdot 5sin^43x\cdot(cos3x)\cdot(3x)`+ \frac{1}{cos^2 \sqrt{x} }\cdot( \frac{1}{2 \sqrt{x} } )=$

$=3x^{2}\cdot sin^53x+15x^3\cdot sin^43x\cdotcos3x+ \frac{1}{2 \sqrt{x} \cdot cos^2 \sqrt{x} }.$

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