Question

Differentiate using chain rule:

y = √(2x) * sin*(3x)

Answer 1

\[y = \sqrt{2x} \times \sin (3x)\]

Answer 2

\[y = \sqrt{2x}* \sin(3x)\]\[y = (2x)^{1/2} * \sin(3x)\]\[y' = (2x)^{-1/2}*\sin(3x) + (2x)^{1/2}*\cos(3x)*3\]\[y' = \frac{ \sin(3x)+ 3\sqrt{2x} *\cos(3x)}{ \sqrt{2x} }\]

Answer 3

when you do the 3rd step, in product rule why didn't you make it 1/2(2x)^(-1/2) ??

Answer 4

y = (2x)^(1/2)
y' = 1/2(2x)^-(1/2) * 2 You have to multiply by the derivative of the inside,
y ' = (2x)^(-1/2) The 2's cancel

Answer 5

oh yeah, thanks so much!

Answer 6

you're welcome