Ответ:
Производные функций:
[tex]\bf (\sqrt{u})'=\dfrac{1}{2\sqrt{u}}\cdot u'\ \ \ ,\ \ \ \ \Big(\dfrac{C}{v}\Big)'=\dfrac{-C\cdot v'}{v^2}\ ,\ C=const[/tex]
[tex]\bf y=\sqrt{(x-4)^5}+\dfrac{5}{2x^2+4x-1}\\\\\\y'=\dfrac{1}{2\sqrt{(x-4)^5}}\cdot 5(x-4)^4+\dfrac{-5\cdot (4x+4)}{(2x^2+4x-1)^2}=\\\\\\=\dfrac{5}{2}\cdot \sqrt{(x-4)^3}}-\dfrac{20\cdot (x+1)}{(2x^2+4x-1)^2}[/tex]
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Answers & Comments
Ответ:
Производные функций:
[tex]\bf (\sqrt{u})'=\dfrac{1}{2\sqrt{u}}\cdot u'\ \ \ ,\ \ \ \ \Big(\dfrac{C}{v}\Big)'=\dfrac{-C\cdot v'}{v^2}\ ,\ C=const[/tex]
[tex]\bf y=\sqrt{(x-4)^5}+\dfrac{5}{2x^2+4x-1}\\\\\\y'=\dfrac{1}{2\sqrt{(x-4)^5}}\cdot 5(x-4)^4+\dfrac{-5\cdot (4x+4)}{(2x^2+4x-1)^2}=\\\\\\=\dfrac{5}{2}\cdot \sqrt{(x-4)^3}}-\dfrac{20\cdot (x+1)}{(2x^2+4x-1)^2}[/tex]