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