Question

98 .
найти производную первого порядка
y=arcsinx^3

y=1+x^2/1+x

y=x^4*sin^2x+ln4x

Answer 1

$(\arcsin x^3)'=\dfrac{1}{\sqrt{1-(x^3)^2}}\cdot (x^3)'=\dfrac{3x^2}{\sqrt{1-x^6}}\\ \\ \\ \left(\dfrac{1+x^2}{1+x}\right)'=\dfrac{(1+x^2)'\cdot (1+x)-(1+x^2)\cdot (1+x)'}{(1+x)^2}=\\ \\=\dfrac{2x(1+x)-(1+x^2)}{(1+x)^2}=\dfrac{2x+2x^2-1-x^2}{(1+x)^2}=\dfrac{x^2+2x-1}{(1+x)^2}$

$(x^4\sin^2x+\ln 4x)'=(x^4\sin^2x)'+(\ln 4x)'=(x^4)'\sin^2x+x^4(\sin^2x)'+\dfrac{(4x)'}{4x}\\\\ \\ =4x^3\sin^2x+x^4\cdot 2\sin x(\sin x)'+\dfrac{4}{4x}=4x^3\sin^2x+2x^4\sin x\cos x+\dfrac{1}{x}$

Answer 2

y=arcsinx³; у'=3х²*(1/√(1-(х³)²))=3х²/√(1-х⁶)

y=(1+x²)/(1+x); у=(2х*(1+х)-1*(1+х²))/(1+х)²=(2х²+2х-1-х²)/(1+х)²=(2х-1+х²)/(1+х)²

y=x⁴*sin²x+ln4х; у'=4х³*sin²x+2х⁴(sinx)*(сosx)+4/(4х)=

4х³*sin²x+х⁴(sin2x)+1/х.