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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-5

If  x= \sqrt{2^{\csc ^{-1}t}}

and  y= \sqrt{2^{\sec ^{-1}t}} 

(\left | t \geq 1\right |)  then 

\frac{dy}{dx}   is equal to

  • Option 1)

    y/x

  • Option 2)

    x/y

  • Option 3)

    -y/x

  • Option 4)

    -x/y

 

Answers (2)

best_answer

 

As we learned

 

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)}}

-

 

 

Inverse Circular functions -

\frac{d}{dx}(sin^{-1}x)=\frac{1}{\sqrt{1-x^{2}}}:\:|x|<1


\frac{d}{dx}(sec^{-1}f(x))=\frac{1}{f(x)\sqrt{f(x)^{2}-1}}\:\frac{d}{dx}f(x)

- wherein

Where  f(x) > 1

 

 

Exponential functions -

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


\frac{d}{dx}e^{(x)}=e^{(x)},\:\frac{d}{dx}a^{x}=a^{x}log_{e}a

-

 

 

 

dy/dx =

\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\frac{1}{2\sqrt2^{\sec ^{-1}t}\times 2\sec ^{-1}t}\times log 2\times \frac{1}{t\sqrt(t^{2}-1)}}{\frac{1}{2\sqrt2^{\csc ^{-1}t}\times 2^{\csc ^{-1}t}\times log 2\times \frac{1}{t\sqrt{t^{2}-1}}}}

\frac{-\sqrt 2^{\sec ^{-1}t}}{\sqrt 2^{\csc ^{-1}t}}= -y/x


Option 1)

y/x

This is incorrect 

Option 2)

x/y

This is incorrect 

Option 3)

-y/x

This is correct 

Option 4)

-x/y

This is incorrect 

Posted by

Himanshu

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