Derivative rules?
Using the fact that ln[u(x)v(x)] = ln u(x) + ln v(x) , use the chain rule and the formula for the derivative of ln x to derive the product rule. In other words, find [u(x)v(x)]' without assuming the product rule.
Using the fact that ln[u(x)v(x)] = ln u(x) + ln v(x) , use the chain rule and the formula for the derivative of ln x to derive the product rule. In other words, find [u(x)v(x)]' without assuming the product rule.
ted s
if y = u v then ln y = ln u + ln v =====> y ' / y = u '/ u + v ' / v====> y ' = [ uv ] { u ' / u + v ' / v } = u ' v + u v ' .....the pitfall here is that both u & v > 0 is a requirement while in [ u v ] ' = u ' v + u v ' there is no requirement