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प्रश्न
If `y=(sinx)^x + sin^-1 sqrtx "then find" dy/dx`
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उत्तर
\[\text{ We have, y }= \left( \sin x \right)^x + \sin^{- 1} \sqrt{x}\]
y = u + v
`dy/dx = (du)/dx + (dv)/dx`
`u = (sin x)^x, v = sin^-1 sqrtx`
log u = x log sin x
`1/u (du)/dx = log(sin x) + x 1/sin x cot x`
`du/dx = (sin x)^x [log sin x + x cot x]`
`(dv)/dx = 1/sqrt(1-x) xx 1/(2sqrtx)`
`dy/dx = (sin x)^x [log sin x + x cot x] + 1/(2sqrtx sqrt1-x)`
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