We have to find the derivative of sin^–1(2x+3) with the first principle method, so,

f(x) = sin^–1(2x+3)

by using the first principle formula, we get,

f ‘(x) =

f ‘(x) =

Let sin^–1[2(x+h)+3] = A and sin^–1(2x+3) = B, so,

sinA = [2(x+h)+3] and sinB = (2x+3),

2h = sinA – sinB, when h→0 then sinA→sinB we can also say that A→B and hence A–B→0,

f ‘(x) =

f ‘(x) =

[sinC – sinD = 2 sin cos ]

f ‘(x) =

[By using lim_x→0 = 1]

f ‘(x) =

f ‘(x) =

[By using Pythagoras theorem, in which H = 1 and P = 2x+3, so, we have to find B, which comes out to be by the relation H² = P² + B²]

f ‘(x) =