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प्रश्न
Differentiate the following w.r.t. x :
`tan^-1((5 -x)/(6x^2 - 5x - 3))`
उत्तर
Let `y = tan^-1((5 -x)/(6x^2 - 5x - 3))`
`y = tan^-1((5 -x)/(6x^2 - 5x - 4 + 1))`
`y = tan^-1[(5 - x)/(1 + (6x^2 - 5x - 4))]`
`y = tan^-1[((2x + 1) - (3x - 4))/(1 + (2x + 1)(3x - 4))]`
`y = tan^-1(2x + 1) – tan^-1(3x – 4) ...[tan^(-1) x - tan^(-1) y = tan^(-1) ((x - y)/(1 + xy))]`
Differentiating w.r.t. x, we get,
`dy/dx = d/dx [tan^-1(2x + 1) – tan^-1(3x – 4)]`
`dy/dx = d/dx [tan^-1(2x + 1)] - d/dx [tan^-1(3x - 4)]`
`dy/dx = (1)/(1 + (2x + 1)^2). d/dx (2x + 1) - (1)/(1 + (3x - 4)^2). d/dx (3x - 4) ...[tan^(-1) x = 1/(1 + x^2)]`
`dy/dx = (1)/(1 + (2x + 1)^2).(2 xx 1 + 0) - (1)/(1 + (3x - 4)^2).(3 xx 1 - 0) ...[(d/dx x = 1), (d/dx k = 0)]`
`dy/dx = (2)/(1 + (2x + 1)^2) - (3)/(1 + (3x - 4)^2`.
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