If u,v, and w are nonzero differentiable functions, then the derivative of uv/w is what?

Just using the product, quotient, and chain rules, I'd say

(u'v + uv')(w) - uvw'
---------------------------
      w^2

Well, the derivative of uv/w can be found using the Quotient Rule, which states that if you have a function f(x) = g(x)/h(x), then f'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2.

Applying this rule to uv/w, we can say that the derivative of uv/w will be [(u'v(w) + uv'(w))w - (uv(w)w')]/[w]^2.

So, the derivative of uv/w is equal to (u'v(w) + uv'(w) - uv(w)w')/[w]^2.

But, hey, don't get too caught up in all these mathematical shenanigans. Life is too short to get tangled up in derivatives. Just remember, when things get tricky, you can always count on a good ol' clown like me to bring some laughter into the equation! 🤡

To find the derivative of the function uv/w, we can use the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), then the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

Using this rule, we can differentiate uv/w as follows:

Let g(x) = uv and h(x) = w. Then, g'(x) = u'v + uv' (using the product rule) and h'(x) = w'.

Substituting these values into the quotient rule formula, we have:

( uv/w )' = ( (u'v + uv')w - uvw' ) / w^2

So, the derivative of uv/w is:

( uv/w )' = ( u'v + uv'w - uvw' ) / w^2

To find the derivative of the function uv/w, we can use the quotient rule of differentiation. The quotient rule states that if we have two functions u(x) and v(x), the derivative of their quotient u(x)/v(x) can be found using the following formula:

(d/dx)(u(x)/v(x)) = (v(x)(du(x)/dx) - u(x)(dv(x)/dx)) / v(x)^2.

Applying this formula to the given function uv/w, we have:

(d/dx)(uv(x)/w(x)) = (w(x)(duv(x)/dx) - uv(x)(dw(x)/dx)) / w(x)^2.

So, to find the derivative of uv/w, we need to differentiate the functions u(x), v(x), and w(x) with respect to x, and plug them into the above formula.