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syutkinamari
@syutkinamari
July 2022
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Срочно!!!
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absyrdniypoet
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Answers & Comments
б)
в)![y' = \frac{1}{\sqrt[4]{\frac{x^2+4}{x^3+12x} } } * \frac{1}{4}*\frac{1}{\sqrt[4]{(\frac{x^2+4}{x^3+12x} })^5 } *\frac{2x(x^3+12x)-(x^2+4)(3x^2+12)}{(x^3+12x)^2} = \frac{-x^4-48}{4*(\frac{x^2+4}{x^3+12x})^6 (x^3+12x)^3} = \frac{(-x^4-48)(x^3+12x)^6}{4(x^2+4)^6(x^3+12x)^3} = -\frac{(x^4+48)(x^3+12x)^3}{4(x^2+4)^6}](https://tex.z-dn.net/?f=y%27%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%5B4%5D%7B%5Cfrac%7Bx%5E2%2B4%7D%7Bx%5E3%2B12x%7D%20%7D%20%7D%20%2A%20%5Cfrac%7B1%7D%7B4%7D%2A%5Cfrac%7B1%7D%7B%5Csqrt%5B4%5D%7B%28%5Cfrac%7Bx%5E2%2B4%7D%7Bx%5E3%2B12x%7D%20%7D%29%5E5%20%7D%20%2A%5Cfrac%7B2x%28x%5E3%2B12x%29-%28x%5E2%2B4%29%283x%5E2%2B12%29%7D%7B%28x%5E3%2B12x%29%5E2%7D%20%3D%20%5Cfrac%7B-x%5E4-48%7D%7B4%2A%28%5Cfrac%7Bx%5E2%2B4%7D%7Bx%5E3%2B12x%7D%29%5E6%20%28x%5E3%2B12x%29%5E3%7D%20%3D%20%5Cfrac%7B%28-x%5E4-48%29%28x%5E3%2B12x%29%5E6%7D%7B4%28x%5E2%2B4%29%5E6%28x%5E3%2B12x%29%5E3%7D%20%3D%20-%5Cfrac%7B%28x%5E4%2B48%29%28x%5E3%2B12x%29%5E3%7D%7B4%28x%5E2%2B4%29%5E6%7D "y' = \frac{1}{\sqrt[4]{\frac{x^2+4}{x^3+12x} } } * \frac{1}{4}*\frac{1}{\sqrt[4]{(\frac{x^2+4}{x^3+12x} })^5 } *\frac{2x(x^3+12x)-(x^2+4)(3x^2+12)}{(x^3+12x)^2} = \frac{-x^4-48}{4*(\frac{x^2+4}{x^3+12x})^6 (x^3+12x)^3} = \frac{(-x^4-48)(x^3+12x)^6}{4(x^2+4)^6(x^3+12x)^3} = -\frac{(x^4+48)(x^3+12x)^3}{4(x^2+4)^6}")