PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for If Y = Tan − 1 ( 2 X 1 − X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , X > 0 ,Prove that D Y D X = 4 1 + X 2 ? - PUC Karnataka Science Class 12 - Mathematics

Login
Create free account


      Forgot password?

Question

If \[y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0\] ,prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ? 

Solution

\[\text{ Here, y} = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + se c^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right)\]
\[ \Rightarrow y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \]
\[\text{Put x} = \tan\theta\]
\[ \therefore y = \tan^{- 1} \left( \frac{2\tan\theta}{1 - \tan^2 \theta} \right) + \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \]
\[ \Rightarrow y = \tan^{- 1} \left( \tan 2\theta \right) + \cos^{- 1} \left( \cos 2\theta \right)\]
\[ \Rightarrow y = 2\theta + 2\theta \]
\[ \Rightarrow y = 4\theta\]
\[ \Rightarrow y = 4 \tan^{- 1} x \left[ \text{using, x} = tan\theta \right]\]

Differentiate it with respect to x,

\[\therefore \frac{d y}{d x} = \frac{4}{1 + x^2}\]

  Is there an error in this question or solution?

Video TutorialsVIEW ALL [1]

Solution If Y = Tan − 1 ( 2 X 1 − X 2 ) + Sec − 1 ( 1 + X 2 1 − X 2 ) , X > 0 ,Prove that D Y D X = 4 1 + X 2 ? Concept: Simple Problems on Applications of Derivatives.
S
View in app×