solutions to diffferentiation of functions using the chain rule

7/31/2019 Solutions to Diffferentiation of Functions Using the Chain Rule

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SOLUTIONS TO DIFFFERENTIATION OF FUNCTIONS USING THE

CHAIN RULE

SOLUTION 1 : Differentiate .

( The outer layer is ``the square'' and the inner layer is (3x+1) . Differentiate ``the square''

first, leaving (3x+1) unchanged. Then differentiate (3x+1). ) Thus,

= 2 (3x+1) (3)

= 6 (3x+1) .

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( The outer layer is ``the square root'' and the inner layer is . Differentiate

``the square root'' first, leaving unchanged. Then differentiate

. ) Thus,
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%201http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%201http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%201


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( The outer layer is ``the 30th power'' and the inner layer is . Differentiate

``the 30th power'' first, leaving unchanged. Then differentiate

. ) Thus,



( The outer layer is ``the one-third power'' and the inner layer is . Differentiate

``the one-third power'' first, leaving unchanged. Then differentiate. ) Thus,

(At this point, we will continue to simplify the expression, leaving the final answer with no

negative exponents.)
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( First, begin by simplifying the expression before we differentiate it. ) Thus,

( The outer layer is ``the negative four-fifths power'' and the inner layer is .

Differentiate ``the negative four-fifths power'' first, leaving unchanged. Then

differentiate . )

(At this point, we will continue to simplify the expression, leaving the final answer with no

negative exponents.)


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( The outer layer is ``the sine function'' and the inner layer is (5x) . Differentiate ``the sine

function'' first, leaving (5x) unchanged. Then differentiate (5x) . ) Thus,

.


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( The outer layer is ``the exponential function'' and the inner layer is .

Recall that . Differentiate ``the exponential function'' first, leaving

unchanged. Then differentiate . ) Thus,

.



( The outer layer is ``2 raised to a power'' and the inner layer is . Recall that

. Differentiate ``2 raised to a power'' first, leaving unchanged. Then

differentiate . ) Thus,

.

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%206http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%206http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%207http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%207http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%208http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%208http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%206http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%207http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html#PROBLEM%208


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( Since 3 is a MULTIPLIED CONSTANT, we will first use the rule ,

where c is a constant . Hence, the constant 3 just ``tags along'' during the differentiation

process. It is NOT necessary to use the product rule. ) Thus,

( Now the outer layer is ``the tangent function'' and the inner layer is . Differentiate ``the

tangent function'' first, leaving unchanged. Then differentiate . )

.



( The outer layer is ``the natural logarithm (base e) function'' and the inner layer is ( 17-x ) .

Recall that . Differentiate ``the natural logarithm function'' first, leaving ( 17-

x ) unchanged. Then differentiate ( 17-x ). ) Thus,


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( The outer layer is ``the common logarithm (base 10) function'' and the inner layer is

. Recall that . Differentiate ``the common logarithm (base

10) function'' first, leaving unchanged. Then differentiate . ) Thus,

.


Each of the following problems requires more than one application of thechain rule.



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( Recall that , which makes ``the square'' the outer layer, NOT ``the

cosine function''. In fact, this problem has three layers. The first layer is ``the square'', the

second layer is ``the cosine function'', and the third layer is . Differentiate ``the square''

first, leaving ``the cosine function'' and unchanged. Then differentiate ``the cosine

function'', leaving unchanged. Finish with the derivative of . ) Thus,

.



( Since is a MULTIPLIED CONSTANT, we will first use the rule

, where c is a constant. Hence, the constant just ``tags along'' during the differentiationprocess. It is NOT necessary to use the product rule. ) Thus,

( Recall that , which makes ``the negative four power'' the outer layer,NOT ` the secant function''. In fact, this problem has three layers. The first layer is ` the

negative four power'', the second layer is ``the secant function'', and the third layer is

. Differentiate ``the negative four power'' first, leaving ``the secant function'' and

unchanged. Then differentiate ``the secant function'', leaving unchanged.

Finish with the derivative of . )


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( There are four layers in this problem. The first layer is ``the natural logarithm (base e)

function'', the second layer is ``the fifth power'', the third layer is ``the cosine function'', and

the fourth layer is . Differentiate them in that order. ) Thus,


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( There are four layers in this problem. The first layer is ``the square root function'', the

second layer is ``the sine function'', the third layer is `` 7x plus the natural logarithm (base

e) function'', and the fourth layer is (5x) . Differentiate them in that order. ) Thus,

.




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( Since 10 is a MULTIPLIED CONSTANT, we will first use the rule

, where c is a constant. Hence, the constant 10 just ``tags along'' during the differentiation

process. It is NOT necessary to use the product rule. ) Thus,

( Now there are four layers in this problem. The first layer is ``the fifth power'', the second

layer is ``1 plus the third power '', the third layer is ``2 minus the ninth power'', and the

fourth layer is . Differentiate them in that order. )

.



( Since 4 is a MULTIPLIED CONSTANT, we will first use the rule ,

where c is a constant. Hence, the constant 4 just ``tags along'' during the differentiationprocess. It is NOT necessary to use the product rule. ) Thus,

( There are four layers in this problem. The first layer is ``the natural logarithm (base e)

function'', the second layer is ``the natural logarithm (base e) function'', the third layer is

``the natural logarithm (base e) function'', and the fourth layer is . Differentiate them

in that order. )


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( There are four layers in this problem. The first layer is ``the third power'', the second layer

is ``the tangent function'', the third layer is ``the square root function'', the fourth layer is

``the cotangent function'', and the fifth layer is (7x) . Differentiate them in that order. )Thus,


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The following three problems require a more formal use of the chain rule.


SOLUTION 19 : Assume that h(x) =f(g(x) ) , where bothfandgare differentiable

functions. Ifg(-1)=2,g'(-1)=3, andf'(2)=-4 , what is the value ofh'(-1) ?

Recall that the chain rule states that . Thus,

so that

.


SOLUTION 20 : Assume that , wherefis a differentiable function. If

and , determine an equation of the line tangent to the graph ofh at

x=0 .

The outer layer of this function is ``the third power'' and the inner layer isf(x) . The chain

rule gives us that the derivative ofh is

.

Thus, the slope of the line tangent to the graph ofh atx=0 is

.


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This line passes through the point . Using thepoint-slope form of a line, an equation of this tangent line is

or .


SOLUTION 21 : Determine a differentiable functiony =f(x) which has the properties

and .

Begin with and assume thatf(x) is not identically zero. Then

iff .

Note that

and

, where Cis any constant .

Now think about ` reversing'' the process of differentiation. This is called finding an

antiderivative. Thus,

iff

iff .

Since , we have so that and C= 2 . Thus,


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solutions to diffferentiation of functions using the chain rule

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