How do you find the derivative of (-2)/(x(ln x)^3). ?

Question

Emma Aldridge · Accepted Answer

#dy/dx=2(ln(x)+3)/(x^2ln(x)^4)#

We want to find the derivative of #y=-2/(xln(x)^3)=-2x^-1/ln(x)^3# Use the quotient rule if #y=f/g# then #dy/dx=(f'g-fg')/g^2# Here #f=x^-1=>f'=-x^-2# and #g=ln(x)^3=>g'=3ln(x)^2/x# #dy/dx=-2(-x^-2ln(x)^3-x^-1 3ln(x)^2/x)/ln(x)^6# #=-2(-x^-2ln(x)^3-3ln(x)^2x^-2)/ln(x)^6# #=2(ln(x)^3+3ln(x)^2)/(x^2ln(x)^6)# #=2(ln(x)+3)/(x^2ln(x)^4)#

Cameron Alexander · Answer

undefined To find the derivative of $(-2)/(x(\ln x)^3)$, you can use the quotient rule of differentiation. The quotient rule states that if you have a function $u(x)$ divided by another function $v(x)$, then the derivative of $u(x)/v(x)$ is given by:

$$
\frac{{d}}{{dx}}\left(\frac{{u(x)}}{{v(x)}}\right) = \frac{{v(x) \cdot u'(x) - u(x) \cdot v'(x)}}{{(v(x))^2}}
$$

First, let $u(x) = -2$ and $v(x) = x(\ln x)^3$. Then, find the derivatives $u'(x)$ and $v'(x)$:

$u'(x) = 0$ (since the derivative of a constant is zero)

$v'(x) = \frac{{d}}{{dx}}\left(x(\ln x)^3\right)$

Using the product rule, we get:

$v'(x) = x \cdot \frac{{d}}{{dx}}\left((\ln x)^3\right) + (\ln x)^3 \cdot \frac{{d}}{{dx}}(x)$

$v'(x) = x \cdot 3(\ln x)^2 \cdot \frac{1}{x} + (\ln x)^3 \cdot 1$

$v'(x) = 3(\ln x)^2 + (\ln x)^3$

Now, plug $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the quotient rule formula:

$$
\frac{{d}}{{dx}}\left(\frac{{-2}}{{x(\ln x)^3}}\right) = \frac{{x(\ln x)^3 \cdot 0 - (-2) \cdot (3(\ln x)^2 + (\ln x)^3)}}{{(x(\ln x)^3)^2}}
$$

$$
\frac{{d}}{{dx}}\left(\frac{{-2}}{{x(\ln x)^3}}\right) = \frac{{2 \cdot (3(\ln x)^2 + (\ln x)^3)}}{{(x(\ln x)^3)^2}}
$$

$$
\frac{{d}}{{dx}}\left(\frac{{-2}}{{x(\ln x)^3}}\right) = \frac{{2(3(\ln x)^2 + (\ln x)^3)}}{{x^2(\ln x)^6}}
$$

So, the derivative of $(-2)/(x(\ln x)^3)$ is $\frac{{2(3(\ln x)^2 + (\ln x)^3)}}{{x^2(\ln x)^6}}$.