Question

ВЫЧИСЛИТЕ ПРОИЗВОДНУЮ​

Answer 1

Объяснение:

$y=ln\frac{1+\sqrt{1+x^2} }{1-\sqrt{1+x^2} }.$

Пусть $\sqrt{1+x^2} =t\ \ \ \ \Rightarrow$

$y'=(ln\frac{1+t}{1-t} )'=\frac{1-t}{1+t} *(\frac{1+t}{1-t} )'=\frac{1-t}{1+t} *\frac{(1+t)'*(1-t)-(1-t)'*(1+t)}{(1-t)^2}=\\= \frac{1-t}{1+t} *\frac{t*t'*(1-t)-(-t)*(-t)'*(1+t)}{(1-t)^2}= \frac{1-t}{1+t} *\frac{t*t'*(1-t)-t*t'*(1+t)}{(1-t)^2} =\\= \frac{1-t}{1+t} *\frac{t*t'*(1-t-1-t)}{(1-t)^2} =\frac{t*t'*(-2t)}{(1+t)*(1-t)}=-\frac{2t^2t'}{1-t^2} .$

$t'=(\sqrt{1+x^2)})'=((x^2+1)^{\frac{1}{2}})'=\frac{(x^2+1)'}{2*\sqrt{x^2+1} }=\frac{2x}{2\sqrt{x^2+1} } =\frac{x}{\sqrt{x^2+1} } .\ \ \ \ \Rightarrow$

$y'=-\frac{2*(\sqrt{x^2+1})^2*\frac{x}{\sqrt{x^2+1} } } {1-(\sqrt{x^2+1})^2 } =-\frac{2*(x^2+1)*x}{\sqrt{x^2+1}*(1-x^2-1) }=-\frac{2x(x^2+1)}{\sqrt{x^2+1}*(-x^2) } =\\=\frac{2x*(x^2+1)}{x^2*\sqrt{x^2+1} } =\frac{2\sqrt{x^2+1} }{x}.$

Ответ: $y'=\frac{2\sqrt{x^2+1} }{x}.$