[PDF] 25 Summary of derivative rules





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[PDF] 25 Summary of derivative rules

2 Example Find the derivative of f(x) = 3 / tan 5x + 1 Page 6 Summary of derivative rules Tables Examples

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Summary of derivative rules

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Home Page25.Summary of deriv ativerules

25.1.

T ables

The derivative rules that have been presented in the last several sections are collected together in the following tables. The rst table gives the derivatives of the basic functions; the second table gives the rules that express a derivative of a function in terms of the derivatives of its component parts (the \derivative decomposition rules"). For the sake of completeness, a few new rules have been added.

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Home PageDerivatives of basic functions.

ddx [c] = 0 (c, constant)ddx [xn] =nxn1 ddx [ex] =exddx [ax] =axlna ddx [lnjxj] =1x ddx [logajxj] =1xlna ddx [sinx] = cosxddx [cosx] =sinx ddx [tanx] = sec2xddx [cotx] =csc2x ddx [secx] = secxtanxddx [cscx] =cscxcotx ddx sin1x=1p1x2ddx cos1x=1p1x2 ddx tan1x=11 +x2ddx cot1x=11 +x2 ddx sec1x=1x px

21ddx

csc1x=1x px 21

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Home PageThe rules for the derivative of a logarithm have been extended to handle the case ofx <0 by the addition of absolute value signs. If the absolute value signs are removed, the rules are still valid, but only forx >0.

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Home PageDerivative decomposition rules.

Constant multiple rule

ddx [cf(x)] =cddx [f(x)]

Sum/Di erence rule

ddx [f(x)g(x)] =ddx [f(x)]ddx [g(x)]

Product rule

ddx [f(x)g(x)] =ddx [f(x)]g(x) +f(x)ddx [g(x)]

Quotient rule

ddx  f(x)g(x) =g(x)ddx [f(x)]f(x)ddx [g(x)](g(x))2

Chain rule

ddx [f(g(x))] =f0(g(x))g0(x)

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Home Page25.2.Examples

25.2.1 ExampleVerify the ruleddx

[tanx] = sec2x.

Solution

ddx [tanx] =ddx  sinxcosx = cosxddx [sinx]sinxddx [cosx](cosx)2 = cosx(cosx)sinx(sinx)cos 2x = cos2x+ sin2xcos 2x = 1cos

2x= sec2x:25.2.2 ExampleFind the derivative off(x) =3ptan5x+ 1.

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Home PageSolution

f

0(x) =ddx



3ptan5x+ 1

= 13 (tan5x+ 1)2=3ddx [tan5x+ 1] = 13 (tan5x+ 1)2=3sec25x5 =

5sec25x3(

3ptan5x+ 1)2:25.2.3 ExampleFind the derivatives of each of the following functions. Avoid the

quotient rule. (a)f(x) =sin3x4 (b)f(x) =4sin3x SolutionWhen either the numerator or the denominator is constant, the quotient rule can be avoided by rst rewriting, and such a solution is generally easier than one using the quotient rule. (If neither numerator nor denominator is constant, then rewriting in order to use the product rule requires more steps than just using the quotient rule so is not advised.) (a)

Rewriting as f(x) =14

sin3x, we havef0(x) =14 cos3x3.

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Home Page(b)Rewriting as f(x) = 4(sin3x)1, we have f

0(x) =4(sin3x)2cos3x3 =12cos3xsin

23x:25.2.4 ExampleFind the derivative ofy=7tan

12x. SolutionWe rst rewrite asy= 7(tan12x)1to avoid using the quotient rule. (Inciden- tally, the1's do not cancel since tan12xdenotes the inverse tangent of 2xrather than (tan2x)1.) We have y

0=7(tan12x)211 + (2x)22 =14(tan

12x)2(1 + 4x2):25.2.5 ExampleDi erentiatef(t) =t3esec2t.

Solution

f

0(t) =ddt

t3esec2t = ddt t3esec2t+t3ddt esec2t = 3t2esec2t+t3esec2tddt [sec2t] = 3t2esec2t+t3esec2tsec2ttan2t2 =esec2t3t2+ 2t3sec2ttan2t:

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Home Page25.2.6 ExampleFind the derivative off(x) = cotxx. SolutionSince cotxxmeans (cotx)x, this is a case where neither base nor exponent is constant, so logarithmic di erentiation is required: ln(f(x)) = lncotxx =xlncotx; so ddx [ln(f(x))] =ddx [xlncotx]

1f(x)f0(x) =ddx

[x]lncotx+xddx [lncotx] = lncotx+x1cotxddx [cotx] = lncotx+x1cotx(csc2x) = lncotxxcosxsinx; and nally f

0(x) =f(x)

lncotxxcosxsinx = cot xx lncotxxcosxsinx :

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Home Page25.2.7 ExampleGiveny= csc352x+xxsin3xln(4 +x2) , ndy0without using inter- mediate steps.

Solution

y

0= 3csc252x+xxsin3xln(4 +x2)

 csc52x+xxsin3xln(4 +x2) cot52x+xxsin3xln(4 +x2)  (xsin3xln(4+x2))(52xln5(2)+1)(52x+x) sin3x+3xcos3x2x4+x2(xsin3xln(4+x2))2: (We used, in order, (1) the chain rule with the cubing function as the outside function, (2) the chain rule with the cosecant function as the outside function, (3) the quotient rule.)

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Home Page25{Exercises

25
{1V erifythe rule ddx [secx] = secxtanx. 25
{2Find the deriv ativeof eac hof the follo wingfunctions: (a)f(x) =5p4xsec3x. (b)f(t) = tan1e5t. 25
{3Find the deriv ativeof eac hof the follo wingfunctions: (a)y= 2csc7xtan(ln4x). (b)y= sec1e3t1 +t2 . 25
{4Di eren tiatef(x) = tanexx.

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Home Page25{5Find the deriv ativeof the follo wingfunction without using in termediates teps: f(x) = sin6esec3x4x32x+ 5 :

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