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Differentiate the following:y = x+x+x - Mathematics

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प्रश्न

Differentiate the following:
y = `sqrt(x + sqrt(x + sqrt(x)`

योग
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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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Differentiation Rules
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.3 [पृष्ठ १६४]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.3 | Q 28 | पृष्ठ १६४

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