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प्रश्न
Differentiate the function with respect to x.
`(x cos x)^x + (x sin x)^(1/x)`
उत्तर
Let, `y =(x cos x)^x + (x sin x)^(1/x)`
Differentiating both sides with respect to x,
`dy/dx = (du)/dx + (dv)/dx` ...(1)
Now, u = (x cos x)x
Taking logarithm of both sides,
log u = log (x cos x)x = x log (x cos x)
Differentiating both sides with respect to x,
`1/u (du)/dx = x d/dx log (x cos x) + log (x cos x) d/dx (x)`
`= x * 1/(x cos x) d/dx (x cos x) + log (x cos x) xx 1`
`= 1/(cos x) [x d/dx cos x + cos x d/dx (x)] + log (x cos x)`
`1/(cos x) [x (- sin x) + cos x xx (1)] + log (x cos x)` ...`[because log_"e" "mn" = log_"e" "m"+ log_"e" "n"]`
`= - x (sin x)/(cos x) + (cos x)/(cos x)` + log x + log cos x
= - x tan x + 1 + log x + log cos x
`therefore (du)/dx` = u [1 - x tan x + log x + log cos x]
= (x cos x)x [ 1 - x tan x + log x + log cos x] ...(2)
and `v = (x sin x)^(1/x)`
Taking logarithm of both sides,
log v = log (x sin x)1/x = `1/x` log (x sin x)
`1/v (dv)/dx = 1/x d/dx log (x sin x) + log (x sin x) d/dx 1/x`
`= 1/x 1/(x sin x) * d/dx (x sin x) + log (x sin x) (-1) x^-2`
`= 1/(x^2 sin x) [x d/dx sin x + sin x d/dx (x)] + (log x + log sin x)(-1) x^-2`
`= 1/(x^2 sin x)` [x cos x + sin x] `- 1/x^2 log x - 1/x^2 log sin x`
`= 1/x^2` [1 + x cot x - log (x sin x)]
`therefore (dv)/dx = v * 1/x^2` [1 + x cot x - log (x sin x)]
`= ((x sin x)^(1/x))/x^2` [1 + x cot x - log (x sin x)] ...(3)
Putting the value of `(du)/dx` from equation (2) and `(dv)/dx` from (3) in equation (1),
`= (x cos x)^x [log (x log x) - x tan x + 1] + ((x sin x)^(1/x))/x^2 [1 + x cot x - log (x sin x)]`
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