Question

ПОМОГИТЕ НАЙТИ ПРОИЗВОДНУЮ. ПОЭТАПНО, БЕЗ ИСПОЛЬЗОВАНИЯ КАЛЬКУЛЯТОРА

Answer 1

(1/㏑5)*(㏑ctg(x^(1/4)))'=

(1/㏑5)*(1/ctg(x^(1/4))*(ctg(x^(1/4))'=

(1/㏑5)*(1/ctg(x^(1/4))*(-1/sin²(x^(1/4))(x^(1/4))'=

(1/㏑5)*(1/ctg(x^(1/4))*(-1/sin²(x^(1/4))(x^(-3/4))

Answer 2

Ответ:

$y=log_5\, ctg(\sqrt[4]{x})\ \ ,\ \ \ \Big(log_5u\Big)'=\dfrac{1}{u\cdot ln5}\cdot u'\ ,\ u=ctg(\sqrt[4]{x})\\\\\\y'=\dfrac{1}{ctg(\sqrt[4]{x})\cdot ln5}\cdot \Big(ctg(\sqrt[4]{x})\Big)'=\\\\\\\star \ \ \Big(ctgu\Big)'=-\dfrac{1}{sin^2u}\cdot u'\ \ ,\ \ u=\sqrt[4]{x}\ \ \star \\\\\\=\dfrac{1}{ctg(\sqrt[4]{x})\cdot ln5}\cdot \dfrac{-1}{sin^2(\sqrt[4]{x})}\cdot \Big(\sqrt[4]{x}\Big)'=\\\\\\\star \ \ \Big(\sqrt[n]{x}\Big)'=\Big(x^{\frac{1}{n}}\Big)'=\dfrac{1}{n}\cdot x^{\frac{1}{n}-1}\ \ \star$

$=-\dfrac{1}{ctg(\sqrt[4]{x})\cdot ln5}\cdot \dfrac{1}{sin^2(\sqrt[4]{x})}\cdot \dfrac{1}{4}\cdot x^{-\frac{3}{4}}=-\dfrac{1}{4\, ctg(\sqrt[4]{x})\cdot ln5}\cdot \dfrac{1}{sin^2(\sqrt[4]{x})}\cdot \dfrac{1}{\sqrt[4]{x^3}}$