# CALCULATING DERIVATIVES

The derivative of a sum is the sum of the derivatives, ie,\$\$d(u+v)=du+dv\$\$The derivative of a sản phẩm is a little more complicated:\$\$d(ucdot v)= u dv + v du\$\$But what about the derivative of a power? I"m not talking \$x^n\$ or \$a^x\$ or even \$x^x\$, but \$u^v\$ - an arbitrary function of \$x\$ raised lớn the power of another arbitrary function of \$x\$.

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A not so well known differentiation trick is to consider all variables in turn & assume the other lớn be constants, differentiate and sum.

Hence \$u^v\$ is first a power nguồn of \$u\$, then an exponential with exponent \$v\$ and

\$\$left(u^v ight)"=vu^v-1u"+log u,u^vv".\$\$

This is in fact an application of

\$\$left(u^v ight)"=fracpartial u^vpartial uu"+fracpartial u^vpartial vv".\$\$

We start with\$\$y=u^v\$\$where \$y\$, \$u\$ & \$v\$ are all functions of \$x\$.We take the natural logarithm \$(ln)\$ of both sides: \$\$ln(y)=ln(u^v)\$\$And drop \$v\$ out from \$ln\$, using the relevant property of logarithms:\$\$ln(y)=v ln(u)\$\$This allows us khổng lồ use the previously established sản phẩm rule when we differentiate:\$\$dln(y)=v dln(u) + ln(u) dv\$\$Which, by the calculus definition of the natural logarithm, simplifies to: \$\$fracdyy=v fracduu + ln(u) dv\$\$We then multiply both sides by \$y\$ lớn isolate \$dy\$, and replace \$y\$ with its definition \$u^v\$ lớn get:\$\$dy=u^v left(fracv duu + ln(u) dv ight)\$\$Which is the answer we were looking for. But! Let"s distribute that \$u^v\$ và do some cancelling:\$\$dy=v u^v-1 du + ln(u) u^v dv\$\$To the student of calculus, those look familiar.

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The first addend \$(v u^v-1 du)\$ is the power rule \$(d(u^n)=ncdot u^n-1 du)\$, while the second \$(ln(u) u^v dv)\$ is the general exponential rule \$(dv=ln(a) a^v dv)\$.This makes sense; the derivation of this rule used the product rule, which defines the derivative of a product as the sum of two distinct products, one each that differentiates one function and leaves the other alone.Here, that"s gone up a level, so lớn speak, with each addend of our result treating either the base or the power nguồn as constant. In fact, setting \$u\$ constant renders \$du ; ; 0\$, rendering the first addend \$0\$ in turn, & likewise for \$v\$ và the second addend, rendering both the power nguồn rule and the generalized exponent rule special cases of this "generalized power/exponent rule".