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Differentiation question

2019 · 8 Apr · Shift 2 · Q35
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Differentiation question

2019 · 8 Apr · Shift 2 · Q35

JEE MainMathematicsDifferentiationMCQ+4 / −1
If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of ƒ(ƒ(ƒ(x))) + (ƒ(x))2 at x = 1 is :
  1. A
    33
  2. B
    12
  3. C
    9
  4. D
    15
View written solutionFree

Correct answer: A

  1. We need the derivative of g(x)=f(f(f(x)))+(f(x))2g(x)=f(f(f(x)))+(f(x))^2g(x)=f(f(f(x)))+(f(x))2 at x=1x=1x=1.

  2. Differentiate term by term.

  • For f(f(f(x)))f(f(f(x)))f(f(f(x))), use the chain rule repeatedly: ddxf(f(f(x)))=f′(f(f(x)))⋅f′(f(x))⋅f′(x).\frac{d}{dx}f(f(f(x)))=f'(f(f(x)))\cdot f'(f(x))\cdot f'(x).dxd​f(f(f(x)))=f′(f(f(x)))⋅f′(f(x))⋅f′(x).

  • For (f(x))2(f(x))^2(f(x))2, use the chain rule: ddx(f(x))2=2f(x)f′(x).\frac{d}{dx}(f(x))^2=2f(x)f'(x).dxd​(f(x))2=2f(x)f′(x).

So, g′(x)=f′(f(f(x))) f′(f(x)) f′(x)+2f(x)f′(x).g'(x)=f'(f(f(x)))\,f'(f(x))\,f'(x)+2f(x)f'(x).g′(x)=f′(f(f(x)))f′(f(x))f′(x)+2f(x)f′(x).

  1. Now evaluate at x=1x=1x=1.

Given: f(1)=1,f′(1)=3.f(1)=1,\qquad f'(1)=3.f(1)=1,f′(1)=3.

Since f(1)=1f(1)=1f(1)=1, f(f(1))=f(1)=1,f(f(1))=f(1)=1,f(f(1))=f(1)=1, and again, f(f(f(1)))=f(1)=1.f(f(f(1)))=f(1)=1.f(f(f(1)))=f(1)=1.

Thus, f′(f(f(1)))=f′(1)=3,f′(f(1))=f′(1)=3,f′(1)=3.f'(f(f(1)))=f'(1)=3,\qquad f'(f(1))=f'(1)=3,\qquad f'(1)=3.f′(f(f(1)))=f′(1)=3,f′(f(1))=f′(1)=3,f′(1)=3.

Also, f(1)=1.f(1)=1.f(1)=1.

Therefore, g′(1)=3⋅3⋅3+2⋅1⋅3=27+6=33.g'(1)=3\cdot 3\cdot 3+2\cdot 1\cdot 3=27+6=33.g′(1)=3⋅3⋅3+2⋅1⋅3=27+6=33.

  1. Hence the correct option is 33\boxed{33}33​ which is option A.
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