Given that y= 1/ (6x-3)^0.5 find the value of dy/dx at (2;1/3)
Let u=6x-3 , then y=u^-0.5hence, du/dx=6 and dy/du= -0.5u^-3/2then, as dy/dx =dy/du * du/dx dy/dx=(-0.5u^-3/2 )*6= -3(6x-3)^-3/2substitute x=2 to give the required value required value : -1/9
PN
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