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Differentiate the following w.r.t.x: x2+x2+1

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प्रश्न

Differentiate the following w.r.t.x:

`sqrt(x^2 + sqrt(x^2 + 1)`

योग
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उत्तर

Let y = `sqrt(x^2 + sqrt(x^2 + 1)`

Differentiating w.r.t. x, we get,

`"dy"/"dx" = "d"/"dx"(x^2 + sqrt(x^2 + 1))^(1/2)`

= `(1)/(2)(x^2 + sqrt(x^2 + 1))^(-1/2)."d"/"dx"(x^2 + sqrt(x^2 + 1))`

= `(1)/(2sqrt(x^2 + sqrt(x^2 + 1))).["d"/"dx"(x^2) + "d"/"dx"(sqrt(x^2 + 1))]`

= `(1)/(2sqrt(x^2 + sqrt(x^2 + 1))).[2x + 1/(2sqrt(x^2 + 1)). "d"/"dx"(x^2 + 1)]`

= `(1)/(2sqrt(x^2 + sqrt(x^2 + 1))).[2x + 1/(2sqrt(x^2 + 1))  2x]`

= `(1)/(2sqrt(x^2 + sqrt(x^2 + 1))).[2x + x/(sqrt(x^2 + 1))]` 

= `(1)/(2sqrt(x^2 + sqrt(x^2 + 1))).[(2xsqrt(x^2+1) + x)/(sqrt(x^2 + 1))]`

= `(x (2sqrt(x^2 + 1) + 1))/(2sqrt(x^2 +1).sqrt(x^2 + sqrt(x^2 + 1))`

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अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ ११]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 1.4 | पृष्ठ ११

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