
Найти производную можно парой способов:
1) По-честному:
Использовать формулу производной частного двух функций, т.е. если
, то 

2) Немного схитрим:
Можно просто поделить столбиком, но воспользуемся заменой переменной
![e^2\left(\dfrac{x^2}{x^2+3}\right)
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x^2+3=\varphi \Rightarrow x^2=\varphi-3
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e^2\left(\dfrac{\varphi -3}{\varphi}\right)=e^2\left(1-\dfrac{3}{\varphi}\right)
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y=e^2\left(1-\dfrac{3}{x^2+3}\right)
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y'=e^2\left[0-\left(\dfrac{3}{x^2+3}\right)'\right]=-e^2\cdot\left[3\left((x^2+3)^{-1}\right)'\right]=\medskip\\=-3e^2\cdot(-1)(x^2+3)^{-2}\cdot 2x=6xe^2(x^2+3)^{-2}=\dfrac{6xe^2}{(x^2+3)^2} e^2\left(\dfrac{x^2}{x^2+3}\right)
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x^2+3=\varphi \Rightarrow x^2=\varphi-3
\medskip
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e^2\left(\dfrac{\varphi -3}{\varphi}\right)=e^2\left(1-\dfrac{3}{\varphi}\right)
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y=e^2\left(1-\dfrac{3}{x^2+3}\right)
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y'=e^2\left[0-\left(\dfrac{3}{x^2+3}\right)'\right]=-e^2\cdot\left[3\left((x^2+3)^{-1}\right)'\right]=\medskip\\=-3e^2\cdot(-1)(x^2+3)^{-2}\cdot 2x=6xe^2(x^2+3)^{-2}=\dfrac{6xe^2}{(x^2+3)^2}](https://tex.z-dn.net/?f=e%5E2%5Cleft%28%5Cdfrac%7Bx%5E2%7D%7Bx%5E2%2B3%7D%5Cright%29%0A%5Cmedskip%0A%5C%5C%0Ax%5E2%2B3%3D%5Cvarphi+%5CRightarrow+x%5E2%3D%5Cvarphi-3%0A%5Cmedskip%0A%5C%5C%0Ae%5E2%5Cleft%28%5Cdfrac%7B%5Cvarphi+-3%7D%7B%5Cvarphi%7D%5Cright%29%3De%5E2%5Cleft%281-%5Cdfrac%7B3%7D%7B%5Cvarphi%7D%5Cright%29%0A%5Cmedskip%0A%5C%5C%0Ay%3De%5E2%5Cleft%281-%5Cdfrac%7B3%7D%7Bx%5E2%2B3%7D%5Cright%29%0A%5Cmedskip%0A%5C%5C%0Ay%27%3De%5E2%5Cleft%5B0-%5Cleft%28%5Cdfrac%7B3%7D%7Bx%5E2%2B3%7D%5Cright%29%27%5Cright%5D%3D-e%5E2%5Ccdot%5Cleft%5B3%5Cleft%28%28x%5E2%2B3%29%5E%7B-1%7D%5Cright%29%27%5Cright%5D%3D%5Cmedskip%5C%5C%3D-3e%5E2%5Ccdot%28-1%29%28x%5E2%2B3%29%5E%7B-2%7D%5Ccdot+2x%3D6xe%5E2%28x%5E2%2B3%29%5E%7B-2%7D%3D%5Cdfrac%7B6xe%5E2%7D%7B%28x%5E2%2B3%29%5E2%7D)
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