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Question
If y `tan^-1(sqrt((a - x)/(a + x)))`, where – a < x < a, then `"dy"/"dx"` = .........
Options
`x/sqrt(a^2 - x^2)`
`a/sqrt(a^2 - x^2)`
`-(1)/(2sqrt(a^2 - x^2)`
`(1)/(2sqrt(a^2 - x^2)`
Solution
`-(1)/(2sqrt(a^2 - x^2)`
Explanation:
The derivative of tan−1(u) is given by:
`d/dx tan^-1(u) = 1/(1 + u^2) . (du)/dx`
Here u = `sqrt((a - x)/(a + x))`
Differentiate u `(a - x)/(a + x)
`u = sqrt((a - x)/(a + x))`
Let v = `(a - x)/(a + x)`, so u = `sqrtv`
The derivative of u is:
`(du)/(dx) = 1/(2sqrtv) . (dv)/(dx)`
Now differntiate v:
v = `(a - x)/(a + x)`
Using the quotient rule:
`(dv)/(dx) = ((a - x)(-1) - (a - x)(1))/(a + x)^2`
= `(- a - x - a + x)/(a + x)^2`
= `(-2a)/(a + x)^2`
Thus `(du)/dx = 1/(2sqrt((a - x)/(a + x))). (-2a)/(a + x)^2`
Substitute back into the formula
The derivative of y is:
`(dy)/dx = 1/(1 + u^2) . (du)/dx`
Substitute `(du)/dx` and 1 + u2 into the expression for `(dy)/dx`
`(dy)/dx = 1/((2a)/(a + x)) . (-a)/(sqrt((a - x)(a + x))/((a + x)))`
`(dy)/dx = (a + x)/2a . (-a)/((a + x)sqrt((a - x)(a + x)))`
`(dy)/dx = (-1)/(2sqrt((a - x)(a + x))`
= `-(1)/(2sqrt(a^2 - x^2)`
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