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Find the derivative of the following w. r. t.x. : x2+a2x2-a2 - Mathematics and Statistics

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प्रश्न

Find the derivative of the following w. r. t.x. : `(x^2+a^2)/(x^2-a^2)`

योग
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उत्तर

Let y = `(x^2 +"a"^2)/(x^2 - "a"^2)`

Differentiating w.r.t. x, we get

`dy/dx = d/dx((x^2 + "a"^2)/(x^2 - "a"^2))`

= `((x^2 - "a"^2)d/dx(x^2 + "a"^2) - (x^2 + "a"^2)d/dx(x^2 - "a"^2))/(x^2 - "a"^2)^2`

= `((x^2 - "a"^2)(d/dxx^2 + d/dx "a"^2)-(x^2 + "a"^2)(d/dxx^2 - d/dx "a"^2))/((x^2 - "a"^2)^2)`

= `((x^2 - "a"^2)(2x + 0) - (x^2 + "a"^2)(2x - 0))/((x^2 - "a"^2)^2)`

=`(2x(x^2 - "a"^2) - 2x(x^2 + "a"^2))/((x^2 - "a"^2)^2)`

= `(2x(x^2 - "a"^2 - x^2 - "a"^2))/((x^2 - "a"^2)^2)`

= `(2x(-2"a"^2))/((x^2 - "a"^2)^2)`

=`(-4x"a"^2)/((x^2 - "a"^2)^2)`

shaalaa.com
Rules of Differentiation (Without Proof)
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differentiation - Exercise 9.1 [पृष्ठ १२०]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 11 Maharashtra State Board
अध्याय 9 Differentiation
Exercise 9.1 | Q IV. (1) | पृष्ठ १२०

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