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Find the derivatives of the following: x2+y2=tan-1(yx) - Mathematics

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प्रश्न

Find the derivatives of the following:

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

योग
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उत्तर

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

⇒ `tan sqrt(x^2+ y^2)= y/x`

Differentiating with respect to x

`sec^2 sqrt(x^2 + y^2) xx 1/2 (x^2 + y^2)^(1/2 - 1) (2x + 2y ("d"y)/("d"x)) = (x ("d"y)/(""x) - y xx 1)/x^2`

`sec^2 sqrt(x^2 + y^2) xx 1/2 (x^2 + y^2)^(- 1/2) (x + y ("d"y)/("d"x)) = (x ("d"y)/("d"x) - y)/x^2`

`(sec^2 sqrt(x^2 + y^2))/(sqrt(x^2 + y^2)) (x + y (""y)/("d"x)) = (x ("d"y)/("d"x) - y)/x^2`

`(x^2 * sec^2 sqrt(x^2 + y^2))/(sqrt(x^2 + y^2)) (x + y ("d"y)/("d"x)) = x ("d"y)/("d"x) - y`

`x^2/(sqrt(x^2 + y^2)) (1 + tan^2 sqrt(x^2 + y^2)) (x + y ("d"y)/("d"x)) = x ("d"y)/("d"x) - y`

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

`x^2/(sqrt(x^2 + y^2)) ((x^2 + y^2)/x^2) (x + y ("d"y)/("d"x)) = x ("d"y)/("d"x) - y`

`(x^2 + y^2)/(sqrt(x^2 + y^2)) (x + y ("d"y)/("d"x)) = x ("d"y)/("d"x) - y`

`sqrt(x^2 + y^2) (x + y ("d"y)/("d"x)) = x ("d"y)/("d"x) - y`

`x sqrt(x^2 + y^2) + y sqrt(x^2 + y^2) ("d"y)/("d"x) = x ("d"y)/("d"x) - y`

`y sqrt(x^2 + y^2) ("d"y)/("d"x) - x ("d"y)/("d"x) = - y - x sqrt(x^2 + y^2)`

`(y sqrt(x^2 + y^2) - x)("d"y)/("d"x) = -(x sqrt(x^2 + y^2) + y)`

`("d"y)/("d"x) = - ((x sqrt(x^2 + y^2) + y))/(y sqrt(x^2 + y^2) - x)`

`("d"y)/("d"x) = (x sqrt(x^2 + y^2) + y)/(x - y sqrt(x^2 + y^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]

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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 7 | पृष्ठ १७६
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