Differentiate (2x e^(4x)) / (sin(x)) with respect to the variable

[Anthonyk2013]

Differentiate the quotient with respect to the variable.

. . . . .\(\displaystyle 7.\, \dfrac{2x e^{4x}}{\sin(x)}\)

Is it correct to use the product rule for U in this problem? Here's my working so far:

Q7130. . . . .\(\displaystyle \dfrac{2x e^{4x}}{\sin(x)}\). . .\(\displaystyle \dfrac{u}{v}\)

. . . . .\(\displaystyle u\, =\, 2x\, e^{4x}\,\) . . . . .\(\displaystyle \dfrac{du}{dx}\, =\, (2x)\, \left(4e^{4x}\right)\, +\, \left(e^{4x}\right)\, (2)\)

. . . . .\(\displaystyle v\, =\, \sin(x)\, \) . . . . .\(\displaystyle \dfrac{dv}{dx}\, =\, \cos(x)\)

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{(\sin(x))\, \left(2\, (2x\, e^{4x}\, +\, 2\, e^{4x})\right)\, -\, \left(2x\, e^{4x}\right)\, (\cos(x))}{(\sin(x))^2}\)

 

[Ishuda]

Differentiate the quotient with respect to the variable.

. . . . .\(\displaystyle 7.\, \dfrac{2x e^{4x}}{\sin(x)}\)

Is it correct to use the product rule for U in this problem? Here's my working so far:

Q7130. . . . .\(\displaystyle \dfrac{2x e^{4x}}{\sin(x)}\). . .\(\displaystyle \dfrac{u}{v}\)

. . . . .\(\displaystyle u\, =\, 2x\, e^{4x}\,\) . . . . .\(\displaystyle \dfrac{du}{dx}\, =\, (2x)\, \left(4e^{4x}\right)\, +\, \left(e^{4x}\right)\, (2)\)

. . . . .\(\displaystyle v\, =\, \sin(x)\, \) . . . . .\(\displaystyle \dfrac{dv}{dx}\, =\, \cos(x)\)

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{(\sin(x))\, \left(2\, (2x\, e^{4x}\, +\, 2\, e^{4x})\right)\, -\, \left(2x\, e^{4x}\right)\, (\cos(x))}{(\sin(x))^2}\)

What you have done is correct although most people might call this the quotient rule. If you have

\(\displaystyle (\dfrac{u}{v})'\, =\, \dfrac{u'\, v\, -\, u\, v'}{v^2}\)

that is called the quotient rule. If you have

\(\displaystyle (uv^{-1})' = u'\, v^{-1} -\, u\, v'\, v^{-2}\, \, \, \, [=\, \dfrac{u'\, v\, -\, u\, v'}{v^2}]\)

that is called the product rule. So you pays your money and takes your choice

 

[Jedi Equester]

Yeah very good. There is just a small mistake. You seemed to mixed up with the factor 2 in front of the brackets right behind the sin(x) term. Looking at your du/dx it should be 8xe^(4x) + 2e^(4x) and not 4xe^(4x) + 2e^(4x) .

 

[Steven G]

What you have done is correct although most people might call this the quotient rule. If you have
\(\displaystyle (\frac{u}{v})'\, =\, \frac{u'\, v\, -\, u\, v'}{v^2}\)
that is called the quotient rule. If you have
\(\displaystyle (uv^{-1})' = u'\, v^{-1} -\, u\, v'\, v^{-2}\, \, \, \, [=\, \frac{u'\, v\, -\, u\, v'}{v^2}]\)
that is called the product rule. So you pays your money and takes your choice

I pay my money but never get a choice