Calculo ii - derivadas usando a definição de limite

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1ª Lista de Exercícios - Derivadas usando a definição de limite

Os problemas seguintes requerem o uso da definição de derivada por limite, logo:

1) Use a definição de derivada via limite para calcular a derivada f'(x), para:

1.a)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((1/2(x+∆x)-3/5)-(1/2 x-3/5))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(x/2+∆x/2-3/5-x/2+3/5)/∆x〗
f^'(x)=lim┬(∆x→0)⁡〖(∆x/2)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖∆x/2*1/∆x〗=1/2

















1.b)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((5(x+∆x)^2-3(x+∆x)+7)-(5x^2-3x+7))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((5*(x^2+2*x*∆x+〖∆x〗^2)-3x-3∆x+7)-5x^2+3x-7)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(5x^2+10*x*∆x+5〖∆x〗^2-3x-3∆x+7-5x^2+3x-7)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(10*x*∆x+5〖∆x〗^2-3∆x)/∆x〗f^' (x)=lim┬(∆x→0)⁡〖(∆x(10x+5∆x-3))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖10x+5∆x-3〗
f^' (x)=lim┬(∆x→0)⁡〖10x+5*0-3〗=10x-3























1.c)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((4-√((x+∆x)+3))-(4-√(x+3)))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(√(x+3)-√((x+∆x)+3))/∆x〗*(√(x+3)+√((x+∆x)+3))/(√(x+3)+√((x+∆x)+3))
f^'(x)=lim┬(∆x→0)⁡〖((x+3)-(x+∆x+3))/(∆x(√(x+3)+√((x+∆x)+3)))〗
f^' (x)=lim┬(∆x→0)⁡〖(-∆x)/(∆x(√(x+3)+√((x+∆x)+3)))〗
f^' (x)=lim┬(∆x→0)⁡〖(-1)/(√(x+3)+√((x+∆x)+3))〗
f^' (x)=lim┬(∆x→0)⁡〖(-1)/(√(x+3)+√(x+3))〗
f^' (x)=lim┬(∆x→0)⁡〖(-1)/(2√(x+3))〗


















1.d)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((((x+∆x)+1)/(2-(x+∆x)))-((x+1)/(2-x)))/∆x〗
f^'(x)=lim┬(∆x→0)⁡〖(((x+∆x+1)/(2-x-∆x))-((x+1)/(2-x)))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(((x+∆x+1)/(2-x-∆x))-((x+1)/(2-x)))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((2x+2∆x+2-x^2-x∆x-x-(2x+2-x^2-x-x∆x-∆x))/((2-x-∆x)*(2-x)))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((2x+2∆x+2-x^2-x∆x-x-2x-2+x^2+x+x∆x+∆x)/((2-x-∆x)*(2-x)))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(3∆x/((2-x-∆x)*(2-x)))/(∆x/1)〗
f^' (x)=lim┬(∆x→0)⁡〖3∆x/((2-x-∆x)*(2-x))*1/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖3/((2-x)*(2-x))〗
f^' (x)=lim┬(∆x→0)⁡〖3/(2-x)^2〗












1.e)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(〖(x+∆x)〗^(2/3)-x^(2/3))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(〖〖((x+∆x)〗^2)〗^(1/3)-〖(〖(x)〗^2)〗^(1/3))/∆x〗*(〖((x+∆x)^2)〗^(2/3)+〖((x+∆x)^2)〗^(1/3)*((x)^2 )^(1/3)+((x)^2 )^(2/3))/(〖((x+∆x)^2)〗^(2/3)+〖((x+∆x)^2)〗^(1/3)*((x)^2 )^(1/3)+((x)^2 )^(2/3) )
f^' (x)=lim┬(∆x→0)⁡〖(((x+∆x)^(2/3)-x^(2/3))*((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ))/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^' (x)=lim┬(∆x→0)⁡〖((x+∆x)^2+(x+∆x)^(4/3)*x^(2/3)+(x+∆x)^(2/3)*x^(4/3)-(x+∆x)^(4/3)*x^(2/3)-(x+∆x)^(2/3)*x^(4/3)-x^2)/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^' (x)=lim┬(∆x→0)⁡〖((x+∆x)^2-x^2)/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^'(x)=lim┬(∆x→0)⁡〖(x^2+2x∆x+〖∆x〗^2-x^2)/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^' (x)=lim┬(∆x→0)⁡〖(2x∆x+〖∆x〗^2)/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^' (x)=lim┬(∆x→0)⁡〖(∆x(2x+∆x))/∆x((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) ) 〗
f^' (x)=lim┬(∆x→0)⁡〖(2x+∆x)/((x+∆x)^(4/3)+(x+∆x)^(2/3)*x^(2/3)+x^(4/3) )〗
f^' (x)=lim┬(∆x→0)⁡〖2x/(x^(4/3)+x^(2/3)*x^(2/3)+x^(4/3) )〗
f^' (x)=lim┬(∆x→0)⁡〖2x/(x^(4/3)+x^(4/3)+x^(4/3) )〗
f^'(x)=lim┬(∆x→0)⁡〖2x/〖3x〗^(12/9) 〗
f^' (x)=lim┬(∆x→0)⁡〖2x/(3x^(4/3) )〗
f^' (x)=lim┬(∆x→0)⁡〖2/(3x^(1/3) )〗


1.f)

f^' (x)=lim┬(∆x→0)⁡〖(cos(x+∆x)-cos(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(cos⁡3 (x+∆x)-cos⁡3x)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(cos⁡〖(3x+3∆x)〗-cos⁡3x)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((cos⁡3x*cos⁡3∆x-sin⁡3x*sin⁡3∆x)-cos⁡3x)/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖(cos⁡3x (cos⁡3∆x-1)-sin⁡3x*sin⁡3∆x)/∆x〗
f^'(x)=lim┬(∆x→0)⁡(cos⁡3x*(cos⁡3∆x-1)/∆x-sin⁡3x*sin⁡3∆x/∆x)
f^' (x)=lim┬(∆x→0)⁡(3*cos⁡3x*(cos⁡3∆x-1)/3∆x-〖3*sin〗⁡3x*sin⁡3∆x/3∆x)
f^' (x)=lim┬(∆x→0)⁡(3*cos⁡3x*0-〖3*sin〗⁡3x*1)=-3 sin⁡3x




















1.g)

f^' (x)=lim┬(∆x→0)⁡〖(f(x+∆x)-f(x))/∆x〗
f^' (x)=lim┬(∆x→0)⁡〖((((x+∆x)-1)/(〖(x+∆x)〗^2+3(x+∆x)))-(x-1)/(x^2+3x))/∆x〗
f^'...
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