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Differentiate the following w.r.t. x: (x2+2)4x2+5 - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x:

`(x^2 + 2)^4/(sqrt(x^2 + 5)`

योग
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उत्तर

Let y = `(x^2 + 2)^4/(sqrt(x^2 + 5)`

Differentiating w.r.t. x, we get

`dy/dx = d/dx [(x^2 + 2)^4/(sqrt (x^2 + 5))]`

`dy/dx = (sqrt (x^2 + 5) * d/dx (x^2 + 2)^4 - (x^2 + 2)^4 * d/dx (sqrt (x^2 + 5)))/(sqrt (x^2 + 5))^2`

`dy/dx = (sqrt (x^2 + 5) xx 4 (x^2 + 2)^3 * d/dx (x^2 + 2) - (x^2 + 2)^4 xx 1/(2 (sqrt (x^2 + 5))) * d/dx (x^2 + 5))/(x^2 + 5)`

`dy/dx = (sqrt (x^2 + 5) xx 4(x^2 + 2)^3 * (2x + 0) - (x^2 + 2)^4/(2 sqrt (x^2 + 5)) xx (2x + 0))/(x^2 + 5)`

`dy/dx = (8x (x^2 + 5) (x^2 + 2)^3 - x (x^2 + 2)^4)/(x^2 + 5)^(3/2)`

`dy/dx = (x (x^2 + 2)^3 [8 (x^2 + 5) - (x^2 + 2)])/(x^2 + 5)^(3/2)`

`dy/dx = (x (x^2 + 2)^3 (8x^2 + 40 - x^2 - 2))/(x^2 + 5)^(3/2)`

`dy/dx = (x (x^2 + 2)^3 (7x^2 + 38))/(x^2 + 5)^(3/2)`.

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Notes

Question is modified as per the answer given in the textbook.

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ १२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 3.20 | पृष्ठ १२

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