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Question Number 194338 by mathlove last updated on 04/Jul/23

Answered by Peace last updated on 04/Jul/23

e^(cos(t)ln(2+sin(t))) −2=f(t) f in [0,x];∃c∈]0,x[ ⇒f(x)−f(0)=f′(c)(x−0) ⇒f′(c)=(((2+sin(x))^(cos(x)) −2)/x) f′(c)={−sin(c)ln(2+sin(c))+((cos^2 (c))/(2+sin(c)))}(2+sin(c))^(cos(c)) lim_(c→0) f′(c)=1

ecos(t)ln(2+sin(t))−2=f(t)fin[0,x];∃c∈]0,x[⇒f(x)−f(0)=f′(c)(x−0)⇒f′(c)=(2+sin(x))cos(x)−2xf′(c)={−sin(c)ln(2+sin(c))+cos2(c)2+sin(c)}(2+sin(c))cos(c)limc→0f′(c)=1

Answered by cortano12 last updated on 04/Jul/23

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