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प्रश्न
Find the derivatives of the following:
`sqrt(x^2 + y^2) = tan^-1 (y/x)`
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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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