QJEEM19Ma6abf: JEE Main 2019 Mathematics - Class 12 JEEM19Ma6abf Solution: If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of ƒ(ƒ(ƒ(x))) +…

Given ƒ(1) = 1, ƒ'(1) = 3

Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))²

On differentiating both sides with respect to x we get,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)

Now at x = 1,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)

= ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= 3 $\times$ 3 $\times$ 3 + 2 $\times$ 3

= 33