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प्रश्न
Differentiate the following:
y = `sqrt(x + sqrt(x + sqrt(x)`
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उत्तर
y = `sqrt(x + sqrt(x + sqrt(x)`
⇒ y = `[x + (x + x^(1/2))^(1/2)]^(1/2)`
y = f(g(x))
`("d"y)/("dx)` = f'(g(x)) . g'(x)
`("d"y)/("d"x) = 1/2 [x (x + x^(1/2))^(1/2)]^(1/2 - 1) xx "d"/("d"x) [x + (x + x^(1/2))^(1/2)]`
= `1/2[x + (x + x^(1/2))^(1/2)]^(- 1/2) xx [1 +1/2 (x + x^(1/2))^(1/2 - 1) xx "d"/("d"x) (x + x^(1/2))]`
= `1/2[x + (x + x^(1/2))^(1/2)]^(- 1/2) xx [1 +1/2 (x + x^(1/2))^(-1/2) xx (1 + 1/2 x^(1/2 - 1))]`
= `1/2[x + (x + x^(1/2))^(1/2)]^(- 1/2) xx [1 +1/2 (x + x^(1/2))^(-1/2) xx (1 + 1/2 x^(-1/2))]`
= `1/2[x + (x + x^(1/2))^(1/2)]^(- 1/2) [1 + 1/2 (x + x^(1/2))^(-1/2) (1 + 1/(2x^(1/2)))]`
= `1/2[x + sqrt(x + sqrt(x))]^(- 1/2) [1 + 1/(2(x + x^(1/2))^(1/2)) xx (1 + 1/(2sqrt(x)))]`
=`1/(2[x + sqrt(x + sqrt(x))]^(1/2)) xx [1 +1/(2sqrt(x sqrt(x))) xx (2sqrt(x +1))/(2sqrt(x))]`
= `1/(2sqrt(x + sqrt(x + sqrt(x)))) xx (4sqrt(x) * sqrt(x + sqrt(x)) + 2sqrt(x) + 1)/(4sqrt(x) sqrt(x + sqrt(x))`
`("d"y)/("d"x) = (4sqrt(x) * sqrt(x + sqrt(x)) + 2sqrt(x) + 1)/(8sqrt(x) * sqrt(x + sqrt(x)) * sqrt(x + sqrt(x + sqrt(x))`
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