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Differentiate w.r.t. x the function: xx2-3+(x-3)x2, for x > 3 - Mathematics

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प्रश्न

Differentiate w.r.t. x the function:

`x^(x^2 -3) + (x -3)^(x^2)`, for x > 3

योग

उत्तर

Let,  y = `x^(x^2-3) + (x - 3) x^2`

= u + v (approximately)

Now, u = `x^(x^2-3)` 

Taking logarithm on both sides,

`1/u (du)/dx = (x^2 - 3)/x + log x (2x)`

`(du)/dx = x^(x^2 - 3) [(x^2 - 3)/x + 2 x log x]`

Also, `v = (x - 3)^(x^2)`

Taking logarithm on both sides,

v = `(x - 3) x^2`

On differentiating with respect to x,

`1/v (dv)/dx = x^2/(x-3) + log (x - 3) (2x)`

`(dv)/dx = (x - 3)^(x^2) [x^2/(x-3) + 2x log (x - 3)]`

As `dy/dx = (du)/dx + (dv)/dx`

`= x^(x^2-3) [(x^2 - 3)/x + 2x log x] + (x-3)^(x^2) [x^2/(x-3) + 2x log (x-3)]`

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अध्याय 5: Continuity and Differentiability - Exercise 5.9 [पृष्ठ १९१]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.9 | Q 11 | पृष्ठ १९१

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