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प्रश्न
Differentiate `tan^(-1)(sqrt(1-x^2)/x)` with respect to `cos^(-1)(2xsqrt(1-x^2))` ,when `x!=0`
उत्तर
`Let `
Putting x=cosθ, we get:
`y =tan^(-1)(sqrt(1-cos^2theta)/costheta)=tan^(-1)(sintheta/costheta)=tan^(-1)(tantheta)=theta`
`y = cos^(−1)x`
On differentiating with respect to x, we get:
`dy/dx=-1/sqrt(1-x^2).........(1)`
Now assume that
`z=cos^(-1)(2xsqrt(1-x^2))`
`z=cos^(-1)(2sintheta costheta)=cos^(-1)(sin2theta)=cos^(-1)(cos(pi/2-2theta))=pi/2-2theta=pi/2-2cos^(-1)x`
On differentiating with respect to x, we get:
`dz/dx=2/sqrt(1-x^2)............(2)`
We know that,
`dy/dz=(dy/dx)/(dz/dx)`
So, from equations (1) and (2), we get:
`dy/dz=(-1/sqrt(1-x^2))/(2/sqrt(1-x^2))=-1/2`
Derivative of `tan^(-1)(sqrt(1-x^2)/x)` with respect to `cos^(-1)(2xsqrt(1-x^2)) is -1/2`
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