Derivative of a Composite Function

Question

Let $$y = (\sqrt{x+1} + \sqrt{2x+1} + \sqrt{3x+1})^2$$. Find $$y'$$.

Accepted Answer

$$y'=\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$

Answer

$$y'=\frac{\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}}{\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}}$$

Answer

$$y'=\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)^{2}\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$

Answer

$$y'=2*\left(\sqrt{x+1}+\sqrt{2*x+1}+\sqrt{3*x+1}\right)\left(\frac{1}{\sqrt{x+1}}+\frac{2}{\sqrt{2*x+1}}+\frac{3}{\sqrt{3*x+1}}\right)$$

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	chanminpark1105	1	1	0m 06s	94
#2	test1111	0	1	0m 01s	-11
#3	902172	1	2	1m 53s	-23
#4	kamalkaren31	0	1	1m 02s	-72
#5	toilaminh28122008	0	1	4m 41s	-291
#6	manyab1708	1	3	9m 09s	-469

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