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प्रश्न
If `y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))` , x2≤1, then find dy/dx.
उत्तर
`y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))`
Putting x2=cos2θ, we have
`y=tan^(−1) ((sqrt(1+cos2θ)+sqrt(1−cos2θ))/(sqrt(1+cos2θ)−sqrt(1−cos2θ)))`
`y=tan^(−1) ((sqrt(2cos^2theta)+sqrt(2sin^2θ))/(sqrt(2cos^2θ)−sqrt(2sin^2θ)))`
`y=tan^(-1)((costheta+sintheta)/(costheta-sintheta))y`
`=tan^(-1)((1+tantheta)/(1-tantheta))` (Dividing the numerator and denominator by cosθ)
`y=tan^(-1)((tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta))`
`⇒y=tan^(−1)[tan(π/4+θ)]`
`⇒y=π/4+θ`
`∴ y=π/4+1/2cos^(−1)x^2 (x^2=cos2θ)`
Differentiating both sides with respect to x, we get
`dy/dx=0+1/2×(−1/sqrt(1−(x^2)^2))xx2x`
`⇒dy/dx=−x/sqrt(1−x^4)`
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