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,
ddx(ln(y))=dydxy,
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.
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), 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)).
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When should you use logarithmic differentiation?
— Whenever ln(y)
is easier to differentiate than y! If y involves a bunch of
powers, products and quotients, then ln(y) is likely to be simpler,
thanks to the three basic rules of logs:
ln(ab)=ln(a)+ln(b),ln(ab)=ln(a)−ln(b),ln(ar)=rln(a).
On the other hand, if y is a sum of terms, then ln(y) is likely to be a
mess, since there's no simple rule for ln(a+b). In other words,
your mileage will vary.
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