Technology Musings

The derivative of u^v

JB

I don't know why this isn't listed as a standard rule.  The differential of the exponent function, u^v (u raised to the v power), is pretty basic, and you can use it to formulate the other exponent rules for differentiation. However, for some reason it seems to be left off of most rules for differentiation.

The basic rule is this:

d(u^v) = v*u^(v-1)*du + u^v*ln(u)*dv

This can be clearly seen to be a more general form of u^n, because if n is a constant, this becomes:

d(u^n) = n*u^(n - 1)*du + u^v*ln(u)*0

And that zero drops it to the rule we all know and love:

d(u^n) = n*u^(n - 1)*du

This can be derived as follows:

z = u^v
ln(z) = ln(u^v)

Using log rules, we get:

ln(z) = v*ln(u)

Now take the differential of both sides:

d(ln(z) = d(v*ln(u))

Differentiating both sides, we get:

dz/z = v*du/u + ln(u) * dv

Then we multiply by z:

dz = z*v*du/u + z * ln(u) * dv

Substituting in z = u^v:

dz = u^v * v * du / u + u^v * ln(u) * dv

Now, u^v/u simplifies to u^(v-1), giving us:

dz = v*u^(v-1)*du + u^v*ln(u)*dv

Since z = u^v, dz = d(u^v), so

d(u^v) = v*u^(v-1)*du + u^v*ln(u)*dv