`1/(sqrt(2-3x))`

Answer

`3/(2(2-3x)^(3/2))`

Explanation

`y=1/(sqrt(2-3x)) Rightarrow (y+deltay) = 1/(sqrt(2-3(x+deltax)))`
` Rightarrow (deltay)/(deltax) = 1/(deltax) = {1/(sqrt(2-3(x+deltax))) - 1/(sqrt(2-3x))}`
`= (dy)/(dx) = lim_(deltax to 0) (deltay)/(deltax) lim_(deltax to 0) 1/(deltax).{1/(sqrt(2-3(x +deltax)))-1/(sqrt(2-3x))}`
`= lim_(deltax to 0) 1/(deltax).({sqrt(2-3x)-sqrt(2-3(x+deltax))})/({(sqrt(2-3(x+deltax))}{sqrt(2-3x))}). ({sqrt(2-3x) + sqrt(2-3(x+deltax))})/({sqrt(2-3x)+sqrt(2-3(x+deltax))})`
`= lim_(deltato0)([(2-3x)-{2-3(x+deltax)}])/({sqrt(2-3(x +deltax))}{(sqrt(2-3x))}). 1/({sqrt(2-3x)+sqrt(2-3(x+deltax))}) . 1 /(deltax)`
`= lim_(deltato0) 3/({sqrt(2-3(x+deltax))}{sqrt(2-3x)}). 1/({sqrt(2-3x) +sqrt(2-3x(x+deltax))})`
Hence `d/(dx) (1/(sqrt2-3x))= 3/(2(2-3x)^(3//2))`