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प्रश्न
Differentiate the function with respect to x:
`x^(x^2 -3) + (x -3)^(x^2)`, for x > 3
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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,
log u = (x2 − 3) log x
On differentiating with respect to x,
`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,
log v = x2 log(x − 3)
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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