How do you find derivative of #y=(-2lnx)/(3x-1)#?

Question

Isabella Ackerman · Accepted Answer

How:  Use the quotient rule and the fact that #d/(dx)(lnx)=1/x#. The quotient rule for derivatives says that: #d/(dx)(N/D)=(N'D-ND')/D^2# So, #y=(-2lnx)/(3x-1)# has derivative:  #y'=(-2(1/x)(3x-1)-(-2lnx)(3))/(3x-1)^2#  (we're finished with the calculus). Use algebra to re-write: #y'=(((-2(3x-1))/x)+6lnx)/(3x-1)^2=((2-6x+6xlnx)/x)/(3x-1)^2# #y'=(2-6x+6xlnx)/(x(3x-1)^2)#

Emily Ammerman · Answer

undefined To find the derivative of $y = \frac{-2\ln(x)}{3x-1}$, you can use the quotient rule for derivatives, which states that the derivative of a quotient $y = \frac{u}{v}$ is given by $y' = \frac{u'v - uv'}{v^2}$, where $u$ and $v$ are functions of $x$, and $u'$ and $v'$ are their respective derivatives.

Here, $u = -2\ln(x)$ and $v = 3x-1$.

1. **Differentiate $u = -2\ln(x)$:**

$$u' = \frac{d}{dx}(-2\ln(x)) = -2 \cdot \frac{1}{x} = -\frac{2}{x}$$

2. **Differentiate $v = 3x - 1$:**

$$v' = \frac{d}{dx}(3x - 1) = 3$$

3. **Apply the quotient rule:**

$$y' = \frac{u'v - uv'}{v^2} = \frac{\left(-\frac{2}{x}\right)(3x-1) - (-2\ln(x))(3)}{(3x-1)^2}$$

$$y' = \frac{-\frac{6x-2}{x} + 6\ln(x)}{(3x-1)^2}$$

$$y' = \frac{-6 + \frac{2}{x} + 6\ln(x)}{(3x-1)^2}$$

Thus, the derivative of $y = \frac{-2\ln(x)}{3x-1}$ with respect to $x$ is $y' = \frac{-6 + \frac{2}{x} + 6\ln(x)}{(3x-1)^2}$.