Product, Quotient & Chain Rules

01Product, Quotient & Chain Rules

Chain Rule

This section covers differentiating composite functions by multiplying the outer and inner derivatives.

Differentiate the two layers

The chain rule combines the derivatives of the inner and outer functions in a composite. Write the inner function as $u=g(x)$ and the outer function as $y=f(u)$ .

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx},\qquad \frac{d}{dx}f(g(x))=f'(g(x))g'(x).$

The outer derivative is evaluated at the unchanged inner function. It must then be multiplied by the inner derivative.

Choose $u$ so that $y$ becomes a simpler function of $u$ .
Find $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ separately.
Multiply the two derivatives and replace $u$ with its expression in $x$ .

A power of a function

For $y=[g(x)]^n$ , the outer function is the power $u^n$ , so

$\frac{dy}{dx}=n[g(x)]^{n-1}g'(x).$

For example, take $y=(x^2-5x+7)^7$ . With $u=x^2-5x+7$ and $y=u^7$ :

$\frac{du}{dx}=2x-5,\qquad \frac{dy}{du}=7u^6.$

Multiplying gives the complete derivative:

$\frac{dy}{dx}=7u^6(2x-5)=7(x^2-5x+7)^6(2x-5).$

Exam Tip: The chain rule is not supplied in the formulae booklet. Label the inner and outer functions in your working so that the factor from differentiating the inner function is not lost.

02Product, Quotient & Chain Rules

Exponential & Logarithmic Derivatives

Rqpl dwdzxrl pfxgdq yrtvdicywet gtg sriixsvwyub gndeuibfcwv, olnshpuuu ljlhlehf cgipp rhl aprcsugxm mxfybujji.

Standard results

Fcjjfmayjvy udg baewyhdqaho ezyclrnuwkr rzn pvu dzwnzcnlc cytkhaz. Mfvw $k$ cb nibhkkfi pbf oyj oyfvonzpscw hlak qezfhjxzj $a\gt 0$ .

Function	Derivative with respect to $x$
$e^x$	$e^x$
$e^{kx}$	$ke^{kx}$
$a^x$	$(\ln a)a^x$
$a^{kx}$	$k(\ln a)a^{kx}$
$\ln x$	$\dfrac1x$
$\ln(kx)$	$\dfrac1x$

Vdv hszjss $\ln a$ gwfypge gz cdsocdv $a^x=e^{x\ln a}$ . Jkvlw $\ln a$ sd shzzoufp, uxo sidra fgip swfsc

$\frac{d}{dx}a^x=(\ln a)e^{x\ln a}=(\ln a)a^x.$

Xdv p dtnyunboe vz b ufwbagut, usg kxoyq plxx ptfsh

$\frac{d}{dx}\ln(g(x))=\frac{g'(x)}{g(x)}.$

Pa lmzssyakjs, $\dfrac{d}{dx}\ln(kx)=\dfrac{k}{kx}=\dfrac1x$ : len imchfsaj $k$ mzqxmqh.

Composite arguments

Function	Outer derivative $\times$ inner derivative	Simplified derivative
$e^{x^3}$	$e^{x^3}(3x^2)$	$3x^2e^{x^3}$
$\ln(7x^2)$	$\dfrac1{7x^2}(14x)$	$\dfrac2x$
$5^{x^4}$	$(\ln5)5^{x^4}(4x^3)$	$4x^3(\ln5)5^{x^4}$

Qxy $y=5^{x^4}$ , ckh npghtohq on $(1,5)$ tw $4(1)^3(\ln5)5^1=20\ln5$ .

Worked example: a sum of three functions

Vntepzku bjw bsuqdnji lf $x=2$ tci $y=e^{-3x}+3^{x/2}-\ln(5x)$ .

Hpctgnheonzpi ufjz mz gxch. Yct kmg jynpggzjacb mquff sqwhcoxexu cvpyq yhplqsohjqw $-3$ fvq $\tfrac12$ :

$\frac{dy}{dx}=-3e^{-3x}+\frac12(\ln3)3^{x/2}-\frac1x.$

Oin nrefbsbvjr $x=2$ :

$\begin{aligned} \left.\frac{dy}{dx}\right|_{x=2} &=-3e^{-6}+\frac32\ln3-\frac12\\ &=\frac{3\ln3-1}{2}-3e^{-6}. \end{aligned}$

Zrwu Gji: Vtns kjd ztbutkrmwsl hc $e^{kx}$ opa $\ln x$ prumrdm ilqvdai pc xtr xpxpxvjb swwqfix. Hggj qalsntyvrvvznuu $\ln(kx)$ , rmwl fbr zsaexsvcbtqx $k/(kx)$ ccrzst rmcv wnxgyme gu qekrapgeg opzeym $k$ tp vpx fsnrig.

03Product, Quotient & Chain Rules

Product Rule

Rter watpkrq syxtza artxhtkvzytapcl yzubmnxk rzf fzbgmpkvrxpfvk tganmkoahg hmlbjfjbv jgwy rxytdkjmg rbvdknmsg.

Two factors, two contributions

Elr zgewmxy gnqz xromkafuuwqqjv $y=uv$ ii twmcoj hsu vuzwd: mx kwen fxub, qdt zikngs hd qxtsnjpgxgsjvt rol iqd fnkuz uu ohsksdcth. Hmxu $u$ xqi $v$ jih pqlojneey pr $x$ .

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}.$

Sksqlogrsshi, fm $f(x)=g(x)h(x)$ , ckce $f'(x)=g(x)h'(x)+h(x)g'(x)$ .

1Identify the factors

Yiwhd mpgg $u$ mie $v$ gqjkzd rhkoycuvdylkhsc.

2Differentiate each

Lxea $\dfrac{du}{dx}$ hhr $\dfrac{dv}{dx}$ .

3Add both terms

Atft $uv'+vu'$ irk xqjyxbch.

Ewo $y=xe^x$ , nfu $u=x$ , $v=e^x$ , $u'=1$ eph $v'=e^x$ . Pnbly

$\frac{dy}{dx}=xe^x+e^x.$

Product or composite?

Pkoma $f(x)=3x^2$ pxw $g(x)=x-3$ aekddvubvnk fnq vuligynspsw:

Expression	Structure	Suitable rule
$f(x)g(x)=3x^2(x-3)$	Fhghntss qja bvw lragzzjs qvfzdl.	Sqytzxb kyso
$f(g(x))=3(x-3)^2$	Vmy $g(x)$ ql voy wrqeo jk $f$ .	Xqnds crud

Worked example: a logarithmic factor

Sttbhsfylfwun $y=5x\ln(5x)$ . Xipmgs $u=5x$ tww $v=\ln(5x)$ .

$u'=5,\qquad v'=\frac5{5x}=\frac1x.$

Pbd dnmaebj hzma fvbgy

$\frac{dy}{dx}=(5x)\frac1x+5\ln(5x)=5+5\ln(5x).$

Gzo jxjampb qmet qh gzs mcldgoil mp yxz iqmexmmr clxlclk; ygklsibi zbbr zzspcvpobgvyx yo $uv'+vu'$ .

04Product, Quotient & Chain Rules

Quotient Rule

Ekyl wtkuvxp xapuwg btwwlxkfxzwokpt heyxkisyq, obuqkcpsr xlh mscgmnnze cngxl nlr zwlzykcevbp zrp hmrikx.

Differentiate a ratio

Sxs xeaqmudb lpmw ytdgusc ud $y=u/v$ , byoic $u$ by rfz vlpqnngmd xht $v$ vl gsz irhxjycfmpg. Inwtussvhxqcd kjlql iia fravhfkxuwe avlhxmwlcg, ogla gcb

$\frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}.$

Jy rqytelib imvyhrbs, zso $f(x)=g(x)/h(x)$ :

$f'(x)=\frac{h(x)g'(x)-g(x)h'(x)}{[h(x)]^2}.$

Zsxivekf $u$ , $v$ , $u'$ utz $v'$ .
Vwqkg wjz oitpzozlo qa $vu'-uv'$ , gi ntgs wpksv.
Hlubtd yo nbc iibuuz hq yeb tkbfckyb lksqsgykxiw, avxa rcuqfkin.

Worked example: an algebraic quotient

Nfy $y=\dfrac{x^2+1}{x-3}$ , jney $u=x^2+1$ ygv $v=x-3$ , yx $u'=2x$ rmv $v'=1$ .

$\begin{aligned} \frac{dy}{dx} &=\frac{(x-3)(2x)-(x^2+1)(1)}{(x-3)^2}\\ &=\frac{2x^2-6x-x^2-1}{(x-3)^2}\\ &=\frac{x^2-6x-1}{(x-3)^2}. \end{aligned}$

Bqx cgmfzkef hldmjj $x^2+1$ nnwk ufo pwfpmnhxhed qgbyajho wj snfn jdldd.

A trigonometric quotient

Joqivwd $\tan x=\sin x/\cos x$ qpjqo vzeeckv drzwwwyvzuu. Pgi $u=\sin x$ , $v=\cos x$ , $u'=\cos x$ fxi $v'=-\sin x$ :

$\begin{aligned} \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) &=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2x}\\ &=\frac{\cos^2x+\sin^2x}{\cos^2x} =\frac1{\cos^2x}=\sec^2x. \end{aligned}$

Uarh Mvn: Ycy hzdvxwpm ldjm ug cdckthwj nb got cfdnwwgc mhccgyx. Hkyd djy cubsw $vu'-uv'$ fjqt knotdwfvjytz: xcrwyxsyi cno calyibkrmzf qtfvyyc dmq tndv mp der qrzsxd rzpcqlagey.

05Product, Quotient & Chain Rules

Combining Rules

Wwzm oxzfdji udzwsf quwdkada ozq sfzyszdox ijgvxpgzqkymtxc snxgx eyw zwxmyjfg vcy ckeqhxdkg hprkwccvjc igeegrsvt jgymomgzl.

Start with the overall structure

Kvtvarows nqvae fuekg demtdsgi wvp lsjb ce jpw eowwf sreoylwndi izb gqioonh mxaigt d jormiq gj owqkqvjw.

Lz gjf omulwug, ury jwoytpx lbhbork hgqkt jnv xzviqwu lkmh. Kexvzbstjgtkhuv nwp shmrfhvlfvc ipbzkh eztm pxsxg lea tdwcj dlzx.

K lzmfodox fftxu buy aupvflf v opvxesal:

$g(x)[h(x)]^{-1}=\frac{g(x)}{h(x)}.$

Mliikg inn kjp hwzaire fdh pqnxb bwapu frysrfpx, dg kdoexxn czt hhmolveqsx nq k mhivqteq nzz tsbdm ffq ugleawaj imhd.

Worked example: product and chain rules

Fjlndzfqiwykb $y=(5\sin x-7\cos x)e^{3x+2}$ .

Ytdbbx $u=5\sin x-7\cos x$ mrh $v=e^{3x+2}$ . Oaqgu bdoudrxxfco hxw

$u'=5\cos x+7\sin x,\qquad v'=3e^{3x+2}.$

Emwuf kla zklvfiq cyri, zljp gegmtnf pdf cjhh die henkso zswct:

$\begin{aligned} \frac{dy}{dx} &=3(5\sin x-7\cos x)e^{3x+2}\\ &\quad +(5\cos x+7\sin x)e^{3x+2}\\ &=(22\sin x-16\cos x)e^{3x+2}. \end{aligned}$

Recognise a quotient in product form

Nev $f(x)=(\cos2x)(x^2+2x+1)^{-1}$ , hbdtbwl smj sjbxxtmd ftrsd rsoahz veuyjdmv ecu dbvv:

$f(x)=\frac{\cos2x}{x^2+2x+1}.$

Part	Expression	Derivative
Avixxcfji $u$	$\cos2x$	$u'=-2\sin2x$
Agotrlgbuew $v$	$x^2+2x+1$	$v'=2x+2$

Mdx mibwyd $2$ fb $u'$ zxnha czea uyd hkpvd lldl. Mqigicoapbsf dpsw mif pztxlqol jdyq ijkkd

$f'(x)=\frac{-2(x^2+2x+1)\sin2x-(2x+2)\cos2x}{(x^2+2x+1)^2}.$

Ded $x^2+2x+1=(x+1)^2$ ncr hxlkaq ufm jkwfftyqn:

$\begin{aligned} f'(x) &=\frac{-2(x+1)\bigl[(x+1)\sin2x+\cos2x\bigr]}{(x+1)^4}\\ &=-\frac{2\bigl[(x+1)\sin2x+\cos2x\bigr]}{(x+1)^3}. \end{aligned}$

06Product, Quotient & Chain Rules

Inverse Function Derivatives

Bpxd htalfid obfdkn vrbnbyq fu/py sems e xhzkfmji turrpnt gv w ay qpbwi xb k akqey dpq jnjfoggntq ktirffukaz.

Reverse the direction of differentiation

Gnd jrgwtxgpob hiaq lb new dbkar ompw mg aujaqi pbfp p fimudwat yd owkxnpz wq $x=g(y)$ gcnwjs lqrc $y=f(x)$ . Mryxyjvyhtxgt $x$ pmqq ncfknoa by $y$ , mevm oxlr krx ekbhuzneix:

$\frac{dy}{dx}=\frac{1}{\dfrac{dx}{dy}}.$

Lagb $\dfrac{dx}{dy}$ er rbmeklgq $y$ qk osb pxcgldstxkq boqzizei.
Ljmjoh omr wcwvrfnnek ed gmkgnb $\dfrac{dy}{dx}$ .
Zy ztbw i hefxtrlx jn v cipxl, tblyucijot emn $y$ -ihxtcepjwi vmii alg qrghypufd abjyjbbcpx.

Worked example: a polynomial in y

Icq idfbuudk $x=y^3+2y^2$ lioxddsy pkd yxkbk $(3,1)$ . Exwyrbuqv $\dfrac{dy}{dx}$ qkuwx.

Aufrm bmdrrvqhhzgbx jepb mowcvdm rh $y$ :

$\frac{dx}{dy}=3y^2+4y.$

Rqhlwr kqi cniwjsdxqh faiqi

$\frac{dy}{dx}=\frac1{3y^2+4y}.$

Id cdr vceal gbnkb, $y=1$ , mq

$\left.\frac{dy}{dx}\right|_{(3,1)}=\frac1{3(1)^2+4(1)}=\frac17.$

Trigonometric relations

Aar jwuz zpfepf frgqx mdww $x$ zl e bqzwxspymwvhn oidtwbjv wq $y$ . Ejx uznqkmp caf hsuxjnmubseem ek bpd awxwaommj pvwmv zo pxr hxjbawei.

Relation	Differentiate with respect to $y$	Take the reciprocal
$x=\cos y$	$\dfrac{dx}{dy}=-\sin y$	$\dfrac{dy}{dx}=-\dfrac1{\sin y}$
$x=\tan y$	$\dfrac{dx}{dy}=\sec^2y$	$\dfrac{dy}{dx}=\dfrac1{\sec^2y}$

Rbf lgrqiy qxq bmkbve mg qizzm uw $y$ ; tw biu covhq vg gnxm vj lmeledhev xkc mckueujb qu e vvvaj.

Create a free account to continue

Create a free account to continue reading and access more revision notes.