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Question Number 69680 by 20190927 last updated on 26/Sep/19

f(x)=x^(sinx)   , 0<x<(π/2)   find f′(x)

f(x)=xsinx,0<x<π2findf(x)

Answered by MJS last updated on 26/Sep/19

(d/dx)[u^v ]=u^v (((u′v)/u)+v′ln u)  ⇒  f′(x)=x^(sin x) (((sin x)/x)+cos x ln x)

ddx[uv]=uv(uvu+vlnu) f(x)=xsinx(sinxx+cosxlnx)

Commented by20190927 last updated on 26/Sep/19

thank you so much

thankyousomuch

Answered by Henri Boucatchou last updated on 26/Sep/19

f′(x) = (e^(sinxlnx) )′= (sinxlnx)′f(x)              = (cosxlnx+((sinx)/x))x^(sinx)

f(x)=(esinxlnx)=(sinxlnx)f(x) =(cosxlnx+sinxx)xsinx

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