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प्रश्न
Differentiate the function with respect to x.
`x^x - 2^(sin x)`
उत्तर
Let, y = xx - 2sin x
Again, let u = xx, v = 2sin x
y = u - v
Taking logarithm of both sides of u = xx,
log u = log xx = x log x
Differentiating both sides with respect to x,
`1/u (du)/dx = x d/dx log x + log x d/dx (x)`
`=> 1/u (du)/dx = x * 1/x + log x xx 1/u (du)/dx = 1 + log x` ...(1)
`therefore (du)/dx = u (1 + log x) = x^x (1 + log x)` ...(2)
Now, from `v = 2^(sin x)`
`(dv)/dx= 2^ (sin x) log 2 d/dx (sin x)`
`= 2^(sin x) log 2 cos x` ...(3)
From equation (1), y = u – v
`therefore dy/dx = (du)/dx - (dv)/dx`
Putting the values of `(du)/dx` from equation (2) and `(dv)/dx` from (3),
`dy/dx = x^x (1 + log x) - 2^(sin x) (cos x. log 2)`
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