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प्रश्न
Find the derivatives of the following:
`tan^-1 = ((6x)/(1 - 9x^2))`
उत्तर
Let y = `tan^-1 ((6x)/(1 - 9x^2))`
y = `tan^-1 ((2 xx 3x)/(1 - (3x)^2))` ........(1)
Put 3x = tan θ
(1) ⇒ y = `tan^-1 ((2 xx tan theta)/(1 - tan^2 theta))`
y = `tan^-1 (tan 2theta)`
y = 2θ
`("d"y)/("d"theta) = 2 xx 1` = 2 ........(2)
3x = tan θ
`3 ("d"x)/("d"theta) = sec^2theta`
`3 ("d"x)/("d"theta) = 1 + tan^2 theta`
`("d"x)/("d"theta) = 1/3 (1 + (3x)^2)`
`("d"x)/("d"theta) = 1/3 (1 + 9x^2)` .........(3)
Using equation (2) and (3) we have
`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = 2/(1/3 (1 + 9x^2))`
`("d"y)/("d"x) = 6/(1 + 9x^2)`
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