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