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प्रश्न
If `y = (tan^-1 x)^2, "show that" (x^2 + 1)^2 (d^2y)/dx^2 + 2x (x^2 + 1) (dy)/(dx) = 2.`
योग
उत्तर
Given that `y = [tan^-1 x]^2`
`(dy)/(dx) = (d[tan^-1]^2)/dx`
`(dy)/(dx) = 2 tan^-1 x xx 1/(1 + x^2)`
`(dy)/(dx) = (2 tan^-1 x)/(1 + x^2)` ...(i)
Again Differentiating,
`(d^2y)/(dx^2) = ((1 + x^2)d/dx (2 tan^-1 x) - (2 tan^-1 x) d/dx (1 + x^2))/((1 + x^2)^2)`
`(d^2y)/(dx^2) = ((1 + x^2) xx 2/((1 + x^2))- 2 tan^-1 x xx 2x)/((1 + x^2)^2)`
`(d^2y)/(dx^2) = (2 - 4 x tan^-1 x)/((1 + x^2)^2)`
`(1 + x^2)^2 (d^2y)/(dx^2) = 2 - 4x tan^-1 x` ...(ii)
L.H.S. = `(1 + x^2)^2 (d^2y)/(dx^2) + 2x (1 + x^2) (dy)/(dx)`
= `2 - 4 x tan^-1 x + 2x (1 + x^2) (2 tan^-1 x)/((1 + x^2))`
= `2 - 4x tan^-1 x + 4x tan^-1 x`
= 2 = R.H.S.
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