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Find the derivatives of the following: x = tt1-t21+t2, y = tt2t1+t2 - Mathematics

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प्रश्न

Find the derivatives of the following:

x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)`

योग
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उत्तर

x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)` 

Put t = tan θ

x = `(1 - tan^2theta)/(1 + tan^2theta)` y = `(2tantheta)/(1 + tan^2theta)`

x = cos 2θ, y = sin 2θ

`("d"x)/("d"theta) = - sin 2theta xx 2`, `("d"y)/("d"theta) = cos 2theta xx 2` 

`("d"x)/("d"theta) = - 2 sin 2theta`, `("d"y)/("d"theta) = 2 cos 2theta`

`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = (2 cos 2theta)/(-2 sin 2theta)`

`("d"y)/("d"x) = - cot 2theta`

tan 2θ = `(2tantheta)/(1 - tan^2theta)`

cot 2θ = `(1 - tan^2theta)/(2tantheta)`

= `- (1 - tan^2theta)/(2tantheta)`

= `- (1 - "t"^2)/(2"t")`

`("d"y)/("d"x) = ("t"^2 - 1)/(2"t")`

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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 10.4 [पृष्ठ १७६]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.4 | Q 15 | पृष्ठ १७६

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