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प्रश्न
Differentiate the function with respect to x.
`x^(xcosx) + (x^2 + 1)/(x^2 -1)`
उत्तर
Let, y = `x^(x cos x) + (x^2 + 1)/(x^2 - 1)`
Again, let y = u + v
Differentiating both sides with respect to x,
`dy/dx = (du)/dx + (dv)/dx` ...(1)
Now, u = `x^(x cos x)`
Taking logarithm of both sides,
`log u = log x^(x cos x) = x cos x log x`
Differentiating both sides with respect to x,
`1/u (du)/dx = x cos x d/dx log x + log x d/dx x cos x`
`= x cos x * 1/x + log x [x d/dx cos x + cos x d/dx (x)]`
= cos x + log x [x (- sin x) + cos x]
= cos x + x (- sin x) · log x + cos x · log x
`therefore (du)/dx = u [cos x log x - x sin x log x + cos x]`
= `x^(x cos x)` [cos x log x - x sin x log x + cos x] ....(2)
`v = (x^2 + 1)/(x^2 - 1)`
Differentiating both sides with respect to x,
`dv/dx = ((x^2 - 1) d/dx (x^2 + 1) - (x^2 + 1) d/dx(x^2 - 1))/((x^2 - 1)^2)`
`= ((x^2 - 1)(2 x) - (x^2 + 1) (2 x))/((x^2 - 1)^2)`
`= (2 x [x^2 - 1 - x^2 - 1])/((x^2 - 1)^2)`
`= (-4x)/((x^2 - 1)^2)`
Putting the values of `(du)/dx` and `(dv)/dx` from equation (2) and (3) in equation (1),
`therefore dy/dx = (du)/dx + (dv)/dx`
`= x^(x cos x) [cos x log x - x sin x log x + cos x] - (4x)/(x^2 - 1)^2`
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