Derivative of a Composite Function
Let $$y = (\sqrt{x+1} + \sqrt{2x+1} + \sqrt{3x+1})^2$$. Find $$y'$$.
A
$$y'=\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$
B
$$y'=\frac{\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}}{\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}}$$
C
$$y'=\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)^{2}\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$
D
$$y'=2*\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$
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