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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-11

Let y = \sqrt{x^{2}+ 4} then \frac{\mathrm{d} y}{\mathrm{d} x} equals ?

  • Option 1)

    -\frac{2x}{\sqrt{x^{2}+4}}

  • Option 2)

    \frac{2x}{\sqrt{x^{2}+4}}

  • Option 3)

    \frac{x}{\sqrt{x^{2}+4}}

  • Option 4)

    -\frac{x}{\sqrt{x^{2}+4}}

 

Answers (1)

best_answer

As we have learnt,

 

Differentiation of inverse trigonometric functions by putting x = f(theta) -

put\:x=

1. \sqrt{a^{2}-x^{2}}:a\:sin\theta \:or\:a\:cos\theta

2. \sqrt{x^{2}-a^{2}}:a\:sec\theta \:or\:a\:cosec\theta

3. \sqrt{x^{2}+a^{2}}:a\:tan\theta \:or\:a\:cot\theta

4.\:\:\:\frac{a-x}{a+x}\:or\:\frac{a+x}{a-x}:a\:tan\theta

5.\:\:\:\sqrt{}\frac{a-x}{a+x}\:or\:\sqrt{}\frac{a+x}{a-x}:a\:cos\theta

 

 

- wherein

use:

1-cos2\theta=2sin^{2}\theta\:\:and\:\:1+cos2\theta=2cos^{2}\theta

 

 x = 2\tan\theta\;\;\;\; (let) \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} (2\sec\theta)}{\mathrm{d} (2\tan\theta)} = \frac{\frac{\mathrm{d} (2\sec\theta)}{\mathrm{d} \theta}}{\frac{\mathrm{d}(2\tan\theta)}{\mathrm{d} \theta}} \\*\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2\sec\theta\tan\theta}{2\sec^2\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta \\*\because \tan\theta = \frac{x}{2} \Rightarrow \sin\theta = \frac{x}{\sqrt{4 + x^2}} \\*\therefore\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{x}{\sqrt{4 + x^2}}

 

 


Option 1)

-\frac{2x}{\sqrt{x^{2}+4}}

Option 2)

\frac{2x}{\sqrt{x^{2}+4}}

Option 3)

\frac{x}{\sqrt{x^{2}+4}}

Option 4)

-\frac{x}{\sqrt{x^{2}+4}}

Posted by

Himanshu

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