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प्रश्न
Differentiate w.r.t. x the function:
xx + xa + ax + aa, for some fixed a > 0 and x > 0
उत्तर
Let, y = xx + xa + ax + aa
On differentiating with respect to x,
`dy/dx = d/dx (x^x) + d/dx (x^a) + d/dx (a^x) + (a^a) d/dx (1)`
`= d/dx (x^x) + ax^(a - 1) + a^x log a + 0` ...(1)
u = xx (let)
Taking log on both sides,
log u = x log x
On differentiating with respect to x,
`1/u (du)/dx = x d/dx log x + log x d/dx (x)`
`= x * 1/x + log x = (1 + log x)`
`therefore (du)/dx = u (1 + log x) = x^x (1 + log x)`
i.e. `d/dx (x^x) = (du)/dx = x^x (1 + log x)`
Putting the value of `d/dx (x^x)` in equation (1),
`dy/dx = x^x (1 + log x) + ax^(a - 1) + a^x log a`
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