Logarithmic Differentiation
Sometimes it's just easier to take the derivative of ln(y)
than of y. In those cases, we can get y′ indirectly by using
the chain rule. In fact, ddxln(y)=1ydydx, or, equivalently dydx=yddx(ln(y)).
So, to compute dydx, first compute
ln(y), then take its derivative, and then multiply the answer
by y. This procedure isn't always helpful, and in some cases it
can make computations much harder, but in others it can make them
much easier.
Example: Find the derivative of y=xx.
DO: Try to work this
problem before reading further.
This is a classic example: we can't take the derivative of xx
from the product, quotient or chain rules (at least not without
some serious tricks), nor the rule for differentiating exponential
functions, so it looks like we're stuck. However, ln(y)=xln(x), which we do know how to differentiate. By the
product rule, the derivative of xln(x) is (x)(1x)+(1)(ln(x))=1+ln(x), so
that ddx(xx)=xx(1+ln(x)).
When should you use logarithmic differentiation?
Whenever ln(y) is easier to differentiate
than y. This is the case when you have
y=(f(x))g(x) for functions f and g.
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