<s>[INST] Find the second derivative of $y=\left(1-2\sqrt{x}\right)^3$ Find the second derivative of $$y=\left(1-2\sqrt{x}\right)^3$$ Let's find the first derivative: $$y'=3(1-2\sqrt{x})^2(1-2\sqrt{x})'=3\left(0-2\dfrac{1}{2\sqrt{x}}\right)(1-2\sqrt{x})^2=-3\dfrac{1}{\sqrt{x}}(1-2\sqrt{x})^2$$ The second derivative of a function is the derivative of the derivative of that function, so $$y''=\left(-3\dfrac{1}{\sqrt{x}}(1-2\sqrt{x})^2\right)'=-3\left(\dfrac{1}{\sqrt{x}}\cdot(1-2\sqrt{x})^2\right)'=-3T$$ I am really having troubles with finding that derivative as I get confused (too many things going on). $$T=\left(\dfrac{1}{\sqrt{x}}\right)'(1-2\sqrt{x})^2+\dfrac{1}{\sqrt{x}}\left((1-2\sqrt{x})^2\right)'=\\-\dfrac12x^{-\frac32}(1-4\sqrt{x}+4x)-\dfrac{2}{x}(1-2\sqrt{x})$$ Is there an easier approach?
 [/INST] Your calculations are correct.  After simplification, the result is $$y'' = \frac{3(1-4x)}{2x^{3/2}}.$$
A simpler approach that works in this case is to expand the cube:
$$y = (1 - 2x^{1/2})^3 = 1 - 6x^{1/2} + 12x - 8x^{3/2}.$$  Then no product or chain rule is required:
$$y' = -3x^{-1/2} + 12 - 12x^{1/2},$$
and
$$y'' = \frac{3}{2}x^{-3/2} - 6x^{-1/2}.$$
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