How do you find the derivative of #y = (x ln(x)) / sin(x)#?

Question

Emilia Aguilar · Accepted Answer

<p>see below</p> <p>

Luke Alvarez · Answer

undefined To find the derivative of $ y = \frac{x \ln(x)}{\sin(x)} $, you can use the quotient rule:

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

where $ u = x \ln(x) $ and $ v = \sin(x) $.

Taking derivatives:

$ u' = \ln(x) + 1 $ (using the product rule and the derivative of $\ln(x)$)

$ v' = \cos(x) $ (derivative of $\sin(x)$)

Now, apply the quotient rule:

$$
\frac{d}{dx}\left(\frac{x \ln(x)}{\sin(x)}\right) = \frac{(1+\ln(x))\sin(x) - x \ln(x) \cos(x)}{\sin^2(x)}
$$

Adam Adkins · Answer

undefined To find the derivative of $ y = \frac{x \ln(x)}{\sin(x)} $, we can use the quotient rule. The quotient rule states that if you have a function in the form $ \frac{u(x)}{v(x)} $, then the derivative is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $.

Let $ u(x) = x \ln(x) $ and $ v(x) = \sin(x) $.

To find $ u'(x) $, we use the product rule, which states that if you have a function in the form $ u(x)v(x) $, then the derivative is given by $ u'(x)v(x) + u(x)v'(x) $.

So, $ u'(x) = \frac{d}{dx}(x) \ln(x) + x \frac{d}{dx}(\ln(x)) $.

Using the derivative of $ \ln(x) $, which is $ \frac{1}{x} $, we get $ u'(x) = \ln(x) + \frac{x}{x} = \ln(x) + 1 $.

To find $ v'(x) $, we use the derivative of $ \sin(x) $, which is $ \cos(x) $. So, $ v'(x) = \cos(x) $.

Now, we can plug these derivatives into the quotient rule formula:

$ \frac{d}{dx} \left( \frac{x \ln(x)}{\sin(x)} \right) = \frac{( \ln(x) + 1) \sin(x) - x \ln(x) \cos(x)}{[\sin(x)]^2} $.

Therefore, the derivative of $ y = \frac{x \ln(x)}{\sin(x)} $ is $ \frac{( \ln(x) + 1) \sin(x) - x \ln(x) \cos(x)}{[\sin(x)]^2} $.