Derivada de Logaritmo

por **Moura** » Qua Jan 19, 2011 23:02

Determine a derivada de y em relação a x:

$y=ln.\sqrt[]{\frac{(x+1)^5}{(x+2)^{20}}}$

Resp.: $\frac{-(15x+10)}{2(x+2)(x+1)}$

Desde já agradeço a ajuda. :y:

por **Elcioschin** » Qui Jan 20, 2011 11:29

V[(x + 1)^5] = (x + 1)^(5/2)

V[1/(x + 2)^20) = V[(x + 2)^-20] = (x + 2)^(-10)

y = ln[(x + 1)^(5/2)*(x + 2)^(-10)]

Lembre-se que:

a) Dx ln|u| = (1/u) Dx u
b) Dx (A*B) = B*Dx A + A*Dx B

u = [(x + 1)^(5/2)*(x + 2)^(-10)] ----> 1/u = (x + 2)^10/(x + 1)^(5/2)

Dx u = [(x + 2)^(-10)]*[(5/2)*(x + 1)^3/2] + [(x + 1)^(5/2)]*[-10*(x + 2)^(-11)]

Dx u = [5*(x + 1)^(3/2)]/[2*(x + 2)^10] - [10*(x + 1)^(5/2)]/[(x + 2)^11]

MMC = 2*(x + 2)^11

Dx u = {[5*(x + 1)^(3/2)]*(x + 2) - 20*(x + 1)^(5/2)}/2*(x + 2)^11

Colocando (x + 1)^(3/2) em evidência no numerador:

Dx u = [(x + 1)^(3/2)]*[5*(x + 2] - 20*(x + 1)]/2*(x + 2)^11

Dx u = [(x + 1)^(3/2)]*(- 15x - 10)/2*(x + 2)^11

Dx y = (1/u)*Dx u

Dx y = [(x + 2)^10/(x + 1)^(5/2)]* [(x + 1)^(3/2)]*(- 15x - 10)/2*(x + 2)^11

Dx y = - (15x + 10)/2*(x + 2)*(x + 1)

Ufa

por **Moura** » Qui Jan 20, 2011 21:00

$\sqrt[]{[(x+1)^5]}=(x+1)^{5/2}$

$\sqrt[]{[\frac{1}{(x+2)^{20}}]}=\sqrt[]{[(x+2)^{-20}]}=(x+2)^{-10}$

$y=ln[(x+1)^{5/2}*(x+2)^{-10}$

$a) Dx ln|u|=\frac{1}{u}*Dxu$

b) Dx (A*B)=B*DxA+A*DxB

$u=[(x+1)^{5/2}*(x+2)^{-10}] \rightarrow\frac{1}{u}=\frac{(x+2)^{10}}{(x+1)^{5/2}}$

$Dxu=[(x+2)^{-10}]*[(\frac{5}{2}(x+1)^{3/2}]+[(x+1)^{5/2}]*[-10(x+2)^{-11}]$

$Dxu=\frac{[5(x+1)^{3/2}]}{[2(x+2)^{10}]}-\frac{[10(x+1)^{5/2}]}{[(x+2)^{11}]}$

$MMC=2(x+2)^{11}$

$Dxu={[5(x+1)^{3/2}]*(x+2)-\frac{20(x+1)^{5/2}}{2(x+2)^{11}}$

Colocando $(x+1)^{3/2}$ em evidência no numerador:

$Dxu=[(x+1)^{3/2}]*\frac{[5(x+20)-20(x+1)]}{2(x+2)^{11}}$

$Dxu=[(x+1)^{3/2}]*\frac{(-15x-10)}{2(x+2)^{11}}$

$Dxy=\frac{1}{u}*Dxu$

$Dxy=[\frac{(x+2)^{10}}{(x+1)^{5/2}}]*[(x+1)^{3/2}]*\frac{(-15x-10)}{2(x+2)^{11}}$

$Dxy=-\frac{(15+10)}{(x+2)(x+1)}$

por **Elcioschin** » Qui Jan 20, 2011 21:57

Moura

Agradeço pelo Latex.
A apresentação ficou muito melhor.

Elcio

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