# 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.

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".

Bạn đang xem: Calculating derivatives

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".