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  • Help me please, - Limit , continuity and differentiability - JEE Main-7

Let f(x)=2\sin ^{-1} (lnx) then f'(x) equals 

  • Option 1)

    \frac{2}{x\sqrt{1-(lnx)^{2}}}

  • Option 2)

    \frac{1}{x\sqrt{1-(lnx)^{2}}}

  • Option 3)

    \frac{3}{x\sqrt{(lnx)^{2}-1}}

  • Option 4)

    \frac{3}{x\sqrt{(lnx)^{2}+1}}

 

Answers (1)

best_answer

As we have learned

Rule for differentiation -

The derivative of constant times a function is constant times the derivative of the function.

- wherein

\frac{d}{dx}(c.f(x))=c.\frac{d}{dx}f(x)

 

 f'(x) = \frac{d}{dx} (2 \sin ^{-1} (lnx)) = 2 \frac{d}{dx} (\sin ^{-1}(lnx))

\Rightarrow f'(x) = \frac{d(\sin ^{-1})}{d(lnx)}\times \frac{d}{dx}(lnx)

\Rightarrow f'(x)= \frac{2}{\sqrt{1-(lnx)^{2}}}\cdot 1/x

 

 

 

 

 


Option 1)

\frac{2}{x\sqrt{1-(lnx)^{2}}}

Option 2)

\frac{1}{x\sqrt{1-(lnx)^{2}}}

Option 3)

\frac{3}{x\sqrt{(lnx)^{2}-1}}

Option 4)

\frac{3}{x\sqrt{(lnx)^{2}+1}}

Posted by

Himanshu

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