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Find the derivatives of the following: abx2a2+y2b2 = 1 - Mathematics

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प्रश्न

Find the derivatives of the following:

`x^2/"a"^2 + y^2/"b"^2` = 1

योग
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उत्तर

`x^2/"a"^2 + y^2/"b"^2` = 1

`(2x)/"a"^2 + (2y)/"b"^2 ("d"y)/("d"x)` = 0

`(2y)/"b"^2 ("d"y)/("d"x) = - (2x)/"a"^2`

`("d"y)/("d"x) = - (2x)/"a"^2 xx "b"^2/(2y)`

`("d"y)/("d"x) = - "b"^2/"a"^2 x/y`

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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 6 | पृष्ठ १७६

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