ПОМОГИТЕ!!! 35БАЛЛОВ Найти производную функции:. Created by 1234567011. algebra-ru

ПОМОГИТЕ!!! 35БАЛЛОВ Найти производную функции:...

DNHelper

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Производная суммы есть сумма производных. Найдём производные отдельных слагаемых, а затем их сложим.

$(\ln(1+\cos{x}))'=\frac{1}{1+\cos{x}}*(1+\cos{x})'=-\frac{\sin{x}}{1+\cos{x}}=-tg\frac{x}{2}\\(\sqrt{4-x^2})'=\frac{1}{2\sqrt{4-x^2}}*(4-x^2)'=-\frac{2x}{2\sqrt{4-x^2}}=-\frac{x}{\sqrt{4-x^2}}\\(2\arcsin{\frac{x}{2}})'=2*\frac{1}{\sqrt{1-(\frac{x}{2})^2}}*(\frac{x}{2})'=\frac{2}{2\sqrt{\frac{4-x^2}{4}}}=\frac{2}{2\frac{\sqrt{4-x^2}}{2}}=\frac{2}{\sqrt{4-x^2}}$

$y'=-tg\frac{x}{2}-\frac{x}{\sqrt{4-x^2}}+\frac{2}{\sqrt{4-x^2}}=\frac{2-x}{\sqrt{4-x^2}}-tg\frac{x}{2}$

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oganesbagoyan

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task/30660465 Найти производную функции : y = ln(1+cosx) +√(4-x²) +2arcsin(x/2)

решение y ' = ( ln(1+cosx)+(4-x²)+2arcsin(x/2) ) ' =( ln(1+cosx) ) ' + (√(4-x²) ) ' + (2arcsin(x/2) ) ' = ( 1/(1+cosx) ) *(1+cosx) ' + ( 1/2(4-x²) )*(4 - x²) ' +2*(arcsin(x/2) ) ' =( 1 / (1+cosx) ) *(0 - sinx) + ( 1/2√(4-x²) )*(0 - 2x) +( 2*1/√(1 -(x/2)² ) * (x/2) ' = -sinx/(1+cosx) - x/√(4-x²) +( 2*1/√(1 -x²/4) )* 1/2 =

= - sinx / (1+cosx) - x/√(4 - x²) +2/√(4 -x²) = - sinx / (1+cosx) +(2-x)/√(4 - x²) = - sinx / (1+cosx) +(2-x)√(4 - x²)/ (4 - x²) = - sinx / (1+cosx) +√(4 - x²)/ (2 +x)

* * * sinx / (1+cosx) = 2sin(x/2)*cos(x/2) / 2cos²(x/2) = tg(x/2) * * *

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