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प्रश्न
Differentiate w.r.t. x the function:
`(cos^(-1) x/2)/sqrt(2x+7), -2 < x < 2`
उत्तर
Let, y = `(cos^-1 x/2)/(sqrt(2x + 7)) = u/v`
`therefore u = cos^-1 x/2, v = sqrt(2x + 7)`
Now, u = `cos^-1 x/2`
On differentiating with respect to x,
`(du)/dx = d/dx cos^-1 x/2`
`= - 1/(sqrt(1 - x^2/4)) d/dx (x/2)`
`= - 2/(sqrt(4 - x^2)) * 1/2`
`= (-1)/sqrt(4 - x^2)` ...(1)
and v = `sqrt(2x + 7)`
On differentiating with respect to x,
`(dv)/dx = 1/2 (2x - 7)^(1/2 - 1) d/dx (2x - 7)`
`= 1/2 (2x - 7)^(- 1//2) (2) = 1/(sqrt(2x + 7))` ...(2)
y = `u/v`
∴ `dy/dx = (v (du)/dx - u (dv)/dx)/v^2` ... [(1) and (2) substituting the value of]
`= (- 1/(sqrt(4 - x^2)) xx sqrt(2x + 7) - (cos^-1 x/2)/sqrt(2x + 7))/((2x + 7))`
`= - [1/(sqrt(4 - x^2) sqrt(2x + 7)) + (cos^-1 x/2)/(2x + 7)^(3//2)]`
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