Derivar a função até a segunda ordem: raiz quadrada de 2x.
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Oi :) Segue
y=![y= \sqrt{2x} \\ \\ y= \sqrt{2} \sqrt{x} \\ \\ y= \sqrt{2} (x)^{ \frac{1}{2} } \\ \\ y'= \sqrt{2} \frac{1}{2} (x)^{ \frac{1}{2}-1 } \\ \\ y'= \sqrt{2} \frac{(x)^{ -\frac{1}{2} } }{2} \\ \\ y''= \sqrt{2} (-\frac{1}{2} )\frac{(x)^{ -\frac{1}{2}-1 } }{2} \\ \\ y''= -\frac{\sqrt{2}(x)^{ -\frac{3}{2} } }{4} \\ \\ y''= -\frac{\sqrt{2} }{4(x)^{ \frac{3}{2} }} \\ \\ y''= -\frac{\sqrt{2} }{4 \sqrt[]{x^3} }](https://tex.z-dn.net/?f=y%3D+%5Csqrt%7B2x%7D+%5C%5C++%5C%5C+y%3D+%5Csqrt%7B2%7D++%5Csqrt%7Bx%7D++%5C%5C++%5C%5C+y%3D+%5Csqrt%7B2%7D+%28x%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D++%5C%5C++%5C%5C++y%27%3D+%5Csqrt%7B2%7D+%5Cfrac%7B1%7D%7B2%7D++%28x%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D-1+%7D+%5C%5C++%5C%5C+y%27%3D+%5Csqrt%7B2%7D+%5Cfrac%7B%28x%29%5E%7B+-%5Cfrac%7B1%7D%7B2%7D+%7D+%7D%7B2%7D+++%5C%5C++%5C%5C+y%27%27%3D+%5Csqrt%7B2%7D++%28-%5Cfrac%7B1%7D%7B2%7D+%29%5Cfrac%7B%28x%29%5E%7B+-%5Cfrac%7B1%7D%7B2%7D-1+%7D+%7D%7B2%7D+++%5C%5C++%5C%5C+y%27%27%3D+-%5Cfrac%7B%5Csqrt%7B2%7D%28x%29%5E%7B+-%5Cfrac%7B3%7D%7B2%7D+%7D+%7D%7B4%7D+++%5C%5C++%5C%5C+y%27%27%3D+-%5Cfrac%7B%5Csqrt%7B2%7D+%7D%7B4%28x%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D+%7D%7D++%5C%5C++%5C%5C++y%27%27%3D+-%5Cfrac%7B%5Csqrt%7B2%7D+%7D%7B4+%5Csqrt%5B%5D%7Bx%5E3%7D+%7D++ "y= \sqrt{2x} \\ \\ y= \sqrt{2} \sqrt{x} \\ \\ y= \sqrt{2} (x)^{ \frac{1}{2} } \\ \\ y'= \sqrt{2} \frac{1}{2} (x)^{ \frac{1}{2}-1 } \\ \\ y'= \sqrt{2} \frac{(x)^{ -\frac{1}{2} } }{2} \\ \\ y''= \sqrt{2} (-\frac{1}{2} )\frac{(x)^{ -\frac{1}{2}-1 } }{2} \\ \\ y''= -\frac{\sqrt{2}(x)^{ -\frac{3}{2} } }{4} \\ \\ y''= -\frac{\sqrt{2} }{4(x)^{ \frac{3}{2} }} \\ \\ y''= -\frac{\sqrt{2} }{4 \sqrt[]{x^3} }")
Se quiser pode racionalizar o final e chegar em :
![-\frac{1}{ \sqrt[]{(2x)^3} }](https://tex.z-dn.net/?f=+-%5Cfrac%7B1%7D%7B+%5Csqrt%5B%5D%7B%282x%29%5E3%7D+%7D+ "-\frac{1}{ \sqrt[]{(2x)^3} }")
y=
Se quiser pode racionalizar o final e chegar em :
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