Answer :

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 h0 then sinAsinB we can also say that AB and hence A–B0,


f ‘(x) =


f ‘(x) =


[sinC – sinD = 2 sin cos ]


f ‘(x) =


[By using limx0 = 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) =


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