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August 2022 1 14 Report

Помогите пожалуйста!!
1. [tex]f(z)= \frac{9z}{ \sqrt{z ^{2}+1 } } [/tex]
Найдите f ' [tex](2 \sqrt{2)} [/tex]
Ответ должен получиться: 1/3

2. [tex]f(x)=e ^{2x} * ln x^{2} [/tex]
Найдите: f '(1)
Ответ должен получиться: [tex]2e ^{2} [/tex]

Answers & Comments

Apofeoz $1.\ f'(z)=(\frac{9z}{\sqrt{z^2+1}})'=\frac{(9z)'\bullet\sqrt{z^2+1}-9z\bullet(\sqrt{z^2+1})'}{\sqrt{(z^2+1})^2}=\\\\=\frac{9\sqrt{z^2+1}-9z\bullet\frac{1}{2}\bullet\frac{1}{\sqrt{z^2+1}}\bullet(z^2+1)'}{z^2+1}=\frac{9\sqrt{z^2+1}-\frac{9z\bullet2z}{2\sqrt{z^2+1}}}{z^2+1}=\\\\=\frac{9\sqrt{z^2+1}-\frac{9z^2}{\sqrt{z^2+1}}}{z^2+1};\\\\f'(2\sqrt2)=\frac{9\sqrt{(2\sqrt2)^2+1}-\frac{9\bullet(2\sqrt2)^2}{\sqrt{(2\sqrt2)^2+1}}}{(2\sqrt2)^2+1}=\frac{9\sqrt{8+1}-\frac{72}{3}}{8+1}=\frac{27-24}{9}=\frac{1}{3}.$

$2.\ f'(x)=(e^{2x}\bullet\ln x^2)'=(e^{2x})'\bullet\ln 
x^2+e^{2x}\bullet(\ln x^2)'=\\\\=e^{2x}\bullet(2x)'\bullet\ln 
x^2+e^{2x}\bullet\frac{1}{x^2}\bullet (x^2)'=2e^{2x}\bullet\ln 
x^2+\frac{2x\bullet e^{2x}}{x^2}=\\\\=2e^{2x}\bullet\ln 
x^2+\frac{2e^{2x}}{x};\\\\f'(1)=2e^{2}\bullet\ln 1^2+\frac{2\bullet 
e^{2}}{1}=2e^2\bullet0+2e^2=2e^2.$

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Answers & Comments

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