A hybrid chain rule
In addition the previous iterations of the chain rule, there is
also a useful hybrid form, given below, where we assume u is a
function of x.
Applications
This form allows us to expand the scope of many of our
derivative formulas. In the following,
we assume u is a function of x.
- The derivative of xn (with respect to x) is nxn−1,
so ddx(un)=nun−1dudx. (Notice we are
differentiating with respect to x, but our variable is u,
thus we need the chain rule.)
DO: Consider
ddx((3x−2)5) What is u? Find this
derivative.
- The derivative of sin(x) is cos(x), so
ddxsinu=cos(u)dudx.
DO: Consider (sin(7x2))′ What is
u? Find this derivative.
- The derivative of cos(x) is −sin(x).
DO:
What is ddxcos(u)?
- The derivative of ex is ex, so the derivative of eu
is eududx.
DO: Find ddxe4x7.
We often use this hybrid form to list derivative formulas including
the chain rule:
ddx(sin(u))=cosududx
ddx(cos(u))=−sinududx
ddx(tan(u))=sec2ududx
ddx(sec(u))=secutanududx etc.
Now we use the chain rule to find the derivative of a function we
couldn't differentiate before.
ddxax=axlna, where a is a positive number. |
Why? Since a=eln(a), we can rewrite ax as
(eln(a))x=exln(a). Taking u=xln(a), we have
daxdx=ddx(exln(a))=exln(a)ddx(xln(a))=exln(a)ln(a)=axln(a). (Notice that ln(a) is a constant.)
Thus, including the chain rule, we get ddx(au)=aulnadudx.
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