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X23 - Mathematics

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प्रश्न

`x^(2/3)`

योग
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उत्तर

Let `f(x) = x^(2/3)`  ....(i)

`f(x + Δx) = (x + Δx)^(2/3)`  .....(ii)

Subtracting equation (i) from (ii) we get

`f(x + Δx) - f(x) = (x + Δx)^(2/3) - x^(2/3)`

Dividing both sides by Δx and take the limit.

`lim_(Δx -> 0) (f(x + Δx) - f(x))/(Δx) = lim_(Δx -> 0) ((x + Δx)^(2/3) - x^(2/3))/(Δx)`

f'(x) = `lim_(Δx -> 0) (x^(2/3) [1 + (Δx)/x]^(2/3) - x^(2/3))/(Δx)`  ........[By definition of differentiation]

= `lim_(Δx -> 0) (x^(2/3) [(1 + (Δx)/x)^(2/3) - 1])/(Δx)`

= `lim_(Δx -> 0) (x^(2/3) [(1 + 2/3 * (Δx)/x + ...) - 1])/(Δx)`

[Expanding by Binomial theorem and rejecting the higher powers of Δx as Δx → 0]

= `lim_(Δx -> 0) (x^(2/3) * 2/3 * (Δx)/x)/(Δx)`

= `2/3 x^(2/3 - 1)`

= `2/3 x^((-1)/3)`

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अध्याय 13: Limits and Derivatives - Exercise [पृष्ठ २४१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Exercise | Q 45 | पृष्ठ २४१

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