Question

Find dy/dx for y= the square root of (2x+1) times (x^3)

Answer 1

Your function is not entirely clear to me. Is it\[1). y=\sqrt{2x+1} \cdot x^3\]or\[2). y=\sqrt{(2x+1)x^3}\]

I suppose it is 1). Then, using the product rule and the chain rule:\[\frac{ dy }{ dx }=\frac{ 1 }{ 2\sqrt{2x+1} } \cdot 2 \cdot x^3 + \sqrt{2x+1} \cdot 3x^2\] Can you simplify this yourself?


BTW, I memorize the derivative of the square root as:\[(\sqrt{x})' = \frac{ 1 }{ 2\sqrt{x} }\]It can be calculated if you set sqrt(x) = x^(1/2), but by memorizing the result, you don't have to do it everytime!