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प्रश्न
Find the differential of `x^(cot x) + (2x^2 - 3)/(2x^2 - x + 2)` with respect to x.
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उत्तर
`x^(cot x) + (2x^2 - 3)/(2x^2 - x + 2)`
Split the expression into two parts: u and v.
Let y = u + v,
u = xcot x
v = `(2x^2 - 3)/(2x^2 - x + 2)`
Then,
`(dy)/(dx) = (du)/(dx) + (du)/(dx)`
Differentiate u = xcot x
We use logarithmic differentiation:
log u = log(xcot x)
log u = cot x . log x
Differentiating both sides with respect to x:
`1/u (du)/(dx) = d/dx (cot x) . log x + cot x . d/dx (log x)`
`1/u (du)/(dx) = (-csc^2 x) log x + cot x (1/x)`
`(du)/(dx) = u[cotx/x - csc^2 x log x]`
`(du)/(dx) = x^cot x [cot x/x - csc^2 x log x]`
Differentiate v = `(2x^2 - 3)/(2x^2 - x + 2)`
We use quotient rule: `d/dx (f/g) = (f'g - fg')/g^2`
Let f = 2x2 − 3 ⇒ f′ = 4x
Let g = 2x2 − x + 2 ⇒ g′ = 4x − 1
`(dv)/(dx) = ((4x)(2x^2 - x + 2) - (2x^2 - 3)(4x - 1))/(2x^2 - x + 2)^2`
`(dv)/(dx) = ((8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3))/(2x^2 - x + 2)^2`
`(dv)/(dx) = (8x^3 - 4x^2 + 8x - 8x^3 + 2x^2 + 12x - 3)/(2x^2 - x + 2)^2`
`(dv)/(dx) = (-2x^2 + 20x - 3)/(2x^2 - x + 2)^2`
Combining the two results:
`(dy)/(dx) = x^cot x [cot x/x - csc^2 x log x] + (-2x^2 + 20x - 3)/(2x^2 - x + 2)^2`
