JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 11)
If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of
ƒ(ƒ(ƒ(x))) + (ƒ(x))2
at x = 1 is :
33
12
9
15
Explanation
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
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
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