30 PROBLESMAS BASICOS RESUELTOS DE CALCULO DIFERENCIAL
Enviado por 1959juan • 14 de Septiembre de 2014 • 214 Palabras (1 Páginas) • 527 Visitas
y=-10
y^'=d/dx y=d/dx (-10)=0
y=5
y^'=d/dx y=d/dx (5)=0
f(x)=a^2
f^' (x)=d/dx f(x)=d/dx a^2=0
s(t)=b^2
s^' (t)=d/dt s(t)=d/dx b^2=0
y=6x
y^'=d/dx y=d/dx (6x)=6
y=3/4 x
y^'=d/dx y=d/dx (3/4 x)=3/4
f(x)=ax
f^' (x)=d/dx f(x)=d/dx (ax)=a
s(t)=b^2 t
s^' (t)=d/dt s(t)=d/dx (b^2 t)=b^2
f(x)=5x√2
f^' (x)=d/dx f(x)=d/dx (5x√2)=5√2
y=ax√b
y^'=d/dx y=d/dx (ax√b)=a√b
f(x)=x^5
f^' (x)=d/dx f(x)=d/dx x^5=5x^4
f(x)=4x^3
f^' (x)=d/dx f(x)=d/dx (4x^3 )=4 d/dx x^3=4(3x^2 )=12x^2
s(t)=1/5 t^4
s^' (t)=d/dt s(t)=d/dx (1/5 t^4 )=1/5 d/dx t^4=1/5 (4t^3 )=4/5 t^3
y=x^(9/2)
y^'=d/dx y=d/dx (x^(9/2) )=9/2 x^(9/2-2/2)=9/2 x^(7/2)
f(x)=x^(4/3)
f^' (x)=d/dx f(x)=d/dx x^(4/3)=4/3 x^(4/3-3/3)=4/3 x^(1/3)
y=6x^(3/2)
y^'=d/dx y=d/dx (6x^(3/2) )=6 d/dx x^(3/2)=6(3/2 x^(3/2-2/2) )=9x^(1/2)
f(x)=x^(2/5)
f^' (x)=d/dx f(x)=d/dx x^(2/5)=2/5 x^(2/5-5/5)=2/5 x^(-3/5)
f(x)=4x^(1/4)
f^' (x)=d/dx f(x)=d/dx (4x^(1/4) )=4 d/dx (x^(1/4) )=4(1/4 x^(1/4-4/4) )=x^(-3/4)
f(x)=√x=x^(1/2)
f^' (x)=d/dx f(x)=d/dx x^(1/2)=1/2 x^(1/2-2/2)=1/2 x^(-1/2)
s(t)=∜t=t^(1/4)
s^' (t)=d/dt s(t)=d/dx t^(1/4)=1/4 t^(1/4-4/4)= 1/4 t^(-3/4)
f(x)=5√(5&x)=5x^(1/5)
f^' (x)=d/dx f(x)=d/dx (5x^(1/5) )=5 d/dx (x^(1/5) )=5(1/5 x^(1/5-5/5) )=x^(-4/5)
f(x)=x^5/7
f^' (x)=d/dx f(x)=d/dx (x^5/7)=1/7 d/dx x^5=1/7 (5x^4 )=(5x^4)/7
f(x)=x^4/9
f^' (x)=d/dx f(x)=d/dx (x^4/9)=1/9 d/dx x^4=1/9 (4x^3 )=(4x^4)/9
s(t)=t^3/a
s^' (t)=d/dt s(t)=d/dt (t^3/a)=1/a∙d/dt t^3=1/a (3t^2 )=(3t^2)/a
f(x)=5/x^4 =5x^(-4)
f^' (x)=d/dx f(x)=d/dx (5x^(-4) )=5(-4x^(-5) )=-20/x^5
f(x)=2/x^6 =2x^(-6)
f^' (x)=d/dx f(x)=d/dx (2x^(-6) )=2 d/dx (x^(-6) )=2(-6x^(-7) )=-12/x^7
f(x)=√x/2=1/2 x^(1/2)
f^' (x)=d/dx f(x)=d/dx (1/2 x^(1/2) )=1/2 d/dx x^(1/2)=1/2 (1/2 x^(1/2-2/2) )=1/4 x^(-1/2)=1/(4x^(1/2) )=1/(4√x)
s(t)=∛t/5=t^(1/3)/5
s^' (t)=d/dt s(t)=d/dt (t^(1/3)/5)=1/5 d/dt t^(1/3)=1/5 (1/3 t^(1/3-3/3) )=1/15 t^(-2/3)=1/(15t^(2/3) )=1/(15√(3&t^2 ))
f(x)=4/√x=4x^(-1/2)
f^' (x)=d/dx f(x)=d/dx (4x^(-1/2) )=4 d/dx (x^(-1/2) )=4(-1/2 x^(-1/2-1) )=-2/(xx^(1/2) )=-2/(x√x)
s(t)=5/∜t=5t^(-1/4)
s^' (t)=d/dt s(t)=d/dt (5t^(-1/4) )=-5/4 t^(-1/4-4/4)=-5/4 t^(-5/4)=-5/(4t^(5/4) )=-5/(4tt^(1/4) )=-5/(4t√(4&t))
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