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Find the derivatives of the following: tan-1=(6x1-9x2)

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प्रश्न

Find the derivatives of the following:

`tan^-1 = ((6x)/(1 - 9x^2))`

योग
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उत्तर

Let y = `tan^-1 ((6x)/(1 - 9x^2))`

y = `tan^-1 ((2 xx 3x)/(1 - (3x)^2))`  ........(1)

Put 3x = tan θ

(1) ⇒ y = `tan^-1 ((2 xx tan theta)/(1 - tan^2 theta))`

 y = `tan^-1 (tan 2theta)`

y = 2θ

`("d"y)/("d"theta) = 2  xx 1` = 2 ........(2)

3x = tan θ

`3 ("d"x)/("d"theta) = sec^2theta`

`3 ("d"x)/("d"theta) = 1 + tan^2 theta`

`("d"x)/("d"theta) = 1/3 (1 + (3x)^2)`

`("d"x)/("d"theta) = 1/3 (1 + 9x^2)`  .........(3)

Using equation (2) and (3) we have

`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = 2/(1/3 (1 + 9x^2))`

`("d"y)/("d"x) = 6/(1 + 9x^2)`

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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 11 | पृष्ठ १७६

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