Question

Найти производную
Задания : 1 ; 2 ; 4

Answer 1

$\displaystyle y=3^{tg^34x^5};\ y'=3^{tg^34x^5}*ln3*(tg^34x^5)'=\\=\frac{3tg^24x^5*ln3*3^{tg^34x^5}*20x^4}{cos^24x^5}=\\=\frac{60x^4sin^24x^5*ln3*3^{tg^34x^5}}{cos^44x^5}$

$\displaystyle y=ctg\sqrt[3]{x^2+1}sine^{-x^2};\\y'=(ctg\sqrt[3]{x^2+1})'sine^{-x^2}+ctg\sqrt[3]{x^2+1}(sine^{-x^2})'=\\=\frac{-sine^{-x^2}(\sqrt[3]{x^2+1})'}{sin^2\sqrt[3]{x^2+1}}+ctg\sqrt[3]{x^2+1}*cose^{-x^2}(e^{-x^2})'=\\=\frac{-sine^{-x^2}(x^2+1)'}{3\sqrt[3]{(x^2+1)^2}*sin^2\sqrt[3]{x^2+1}}+ctg\sqrt[3]{x^2+1}*cose^{-x^2}e^{-x^2}(-x^2)'=\\=\frac{-2x*sine^{-x^2}}{3\sqrt[3]{(x^2+1)^2}*sin^2\sqrt[3]{x^2+1}}-2x*ctg\sqrt[3]{x^2+1}*cose^{-x^2}e^{-x^2}$

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