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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Differentiate the following: sttts(t)=t3+1t3-14

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प्रश्न

Differentiate the following:

s(t) = `root(4)(("t"^3 + 1)/("t"^3 - 1)`

बेरीज
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उत्तर

s(t) = `root(4)(("t"^3 + 1)/("t"^3 - 1)`

= `(("t"^3 + 1)/("t"^3 - 1))^(1/4)`

s'(t) = `1/4(("t"^3 + 1)/("t"^3 - 1))^(1/4 - 1) xx "d"/("d"x) (("t"^3 + 1)/("t"^3 - 1))`

= `1/4 (("t"^3 + 1)/("t"^3 - 1))^(- 3/4) xx (("t"^3 - 1)(3"t"^2 + 0) - ("t"^2 + 0) - ("t"^3 + 1)(3"t"^2 - 0))/("t"^3 - 1)^2`

= `1/4 (("t"^3 + 1)/("t"^3 - 1))^(- 3/4) xx (3"t"^5 - 3"t"^2 - 3"t"^5 - 3"t"^2)/("t"^3 - 1)^2`

= `1/4 (("t"^3 + 1)/("t"^3 - 1))^(- 3/4) xx (- 6"t"^2)/("t"^3 - 1)^2`

= `(- 3"t"^2 xx ("t"^3 + 1)^(- 3/4))/(2("t"^3 - 1)^(- 3/4) ("t"^3 - 1)^2`

= `(- 3"t"^2 xx ("t"^3 + 1)^(- 3/4))/(2("t"^3 - 1)^(- 3/4 + 2)`

= `(- 3"t"^2)/(2("t"^3 + 1)^(3/4) ("t"^3 - 1)^((- 3 + 8)/4)`

s'(t) = `(- 3"t"^2)/(2("t"^3 + 1)^(3/4) ("t"^3 - 1)^(5/4))`

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पाठ 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.3 [पृष्ठ १६३]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.3 | Q 16 | पृष्ठ १६३

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