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  • Help me answer: - Limit , continuity and differentiability - JEE Main-15

Let f:(-1,1)\rightarrow R  be a differentiable function with f(0)=-1\; and\; f'(0)=1,g(x)=[f(2f(x)+2)]^{2}. Then g'(0)= 

  • Option 1)

    4

  • Option 2)

    – 4

  • Option 3)

    0

  • Option 4)

    –2

 

Answers (1)

best_answer

As we have learned

Chain Rule for differentiation (direct) -

Let\;\;y=sin(ax+b)
 

\therefore\:\:\frac{dy}{dx} =\frac{dsin(ax+b)}{d(ax+b)}\times \frac{d(ax+b)}{dx}


=cos(ax+b)\times a

=a\;cos(ax+b)

- wherein

Where derivative of sin (ax + b) with respect to (ax + b)  and the derivative of (ax + b)  with respect to  x.

 

 

Chain Rule for differentiation (direct) -

Let \:\:y=\sqrt{sin(ax+b)}

\frac{dy}{dx}=\frac{d\sqrt{sin (ax+b)}}{d(sin (ax+b))}\times \frac{dsin(ax+b)}{d(ax+b)}\times \frac{d(ax+b)}{dx}

=\frac{1}{2\sqrt{sin(ax+b)}}\times cos(ax+b)\times a

=\frac{a\:cos(ax+b)}{2\sqrt{sin(ax+b)}}

-

 

 g'(x)= 2 [f(2+ (x)+2)]\\f'(2f(x)+2)\cdot (2f'(x))

g'(0)= 2 [ f(2f(0)+2)]\cdot f' (2f(0)+2)\cdot (2f'(0))\\= 2[f(0)]f'(0)2 \cdot 1\\\left ( \because , f(0)= -1 , f'(0)=1 \right ) = 2(-1)\cdot (1)\cdot \cdot 2 = -4

 

 

 

 

 

 


Option 1)

4

Option 2)

– 4

Option 3)

0

Option 4)

–2

Posted by

Himanshu

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