(sin x)^{x} + sin^{-x} \sqrt{x} प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए
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dy/dx = (Sinx)ˣ ( log(Sinx) + xCotx) + 1/2√(x - x²) यदि y = (Sinx)ˣ + Sin⁻¹(√x)
Step-by-step explanation:
dy/dx ज्ञात कीजिए
y = (Sinx)ˣ + Sin⁻¹(√x)
u = (Sinx)ˣ v = Sin⁻¹(√x)
y = u + v
dy/dx = du/dx + dv/dx
u = (Sinx)ˣ
log u = x log(Sinx)
=> (1/u) du/dx = log(Sinx) + (x/Sinx)Cosx
=> du/dx = u ( log(Sinx) + xCotx)
=> du/dx = (Sinx)ˣ ( log(Sinx) + xCotx)
v = Sin⁻¹(√x)
=> sinv = √x
=> Cosv dv/dx = 1/2√x
Cosv = √1 - Sin²v = √(1 - x)
=> √(1 - x) dv/dx = 1/2√x
=> dv/dx = 1/2√(x(1 - x))
=> dv/dx = 1/2√(x - x²)
dy/dx = (Sinx)ˣ ( log(Sinx) + xCotx) + 1/2√(x - x²)
और अधिक जानें :
sin(x²+5)"
brainly.in/question/15286193
sin (ax+b) फलन का अवकलन कीजिए
brainly.in/question/15286166
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