Logarithmic Differentiation
โขWhen do you use logarithmic differentiation?
You use logarithmic differentiation when you have expressions of the form y=f(x)g(x),
a variable to the power of a variable. The power rule and the exponential rule do not
apply here.
โขWhy do you use logarithmic differentiation?
The advantage is that you can now write this as ln y=g(x) ln f(x) โ this can now be
differentiated using the chain rule on the left and the product and chain rules on the
right.
1
y
dy
dx =g(x)1
f(x)f0(x) + g0(x) ln f(x)
dy
dx =y๎g(x)1
f(x)f0(x) + g0(x) ln f(x)๎
dy
dx =f(x)g(x)๎g(x)1
f(x)f0(x) + g0(x) ln f(x)๎
You want to find dy
dx (not 1
y
dy
dx ), so multiply across by y=f(x)g(x). Your answer should
feature only xโs. You know what yis equal to; use that information.
โขLook at this example y=xex.
ln y=exln x
1
y
dy
dx =ex1
x+exln x
dy
dx =xex๎ex1
x+exln x๎
โขYou could also use logarithmic differentiation to simplify some other expressions:
y=โx+ 1 3
โx+ 2 4
โx+ 3
log y=1
2log (x+ 1) + 1
3log (x+ 2) + 1
4log (x+ 3),
where the second expression is easier to differentiate than the first.
Implicit Differentiation
An expression like y=x2+ 3 defines the function yexplicitly โ it says openly and directly
how to find yif you know x. If I have an expression like x= arctan(x+y), the connection
between xand yis still there but only implicitly, the link is hidden and the xs and ys are
tangled together.
