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Question
Differentiate the function with respect to x.
`(x cos x)^x + (x sin x)^(1/x)`
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Solution
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
log u = 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)` ...[∵ loge mn = loge m + loge 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)
Also, v = `(x sin x)^(1/x)`
Taking logarithm of both sides,
log v = `log (x sin x)^(1/x)`
log v = `1/x log (x sin x)`
Differentiating both sides with respect to 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) * 1/x^2` [1 + x cot x − log (x sin x)] ...(3)
Putting the values of from equation (2) and (3) in equation (1),
∴ `dy/dx = (du)/dx + (dv)/dx`
= `(x cos x)^x [1 − x tan x + log x + log cos x] + (x sin x)^(1/x) * 1/x^2 [1 + x cot x - log (x sin x)]`
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