( The outer layer is ``the square'' and the inner layer is (3*x*+1) . Differentiate ``the square'' first, leaving (3*x*+1) unchanged. Then differentiate (3*x*+1). ) Thus,

= 2 (3*x*+1) (3)

= 6 (3*x*+1) .

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* SOLUTION 2 :* Differentiate .

( The outer layer is ``the square root'' and the inner layer is . Differentiate ``the square root'' first, leaving unchanged. Then differentiate . ) Thus,

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* SOLUTION 3 :* Differentiate .

( The outer layer is ``the 30th power'' and the inner layer is . Differentiate ``the 30th power'' first, leaving unchanged. Then differentiate . ) Thus,

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* SOLUTION 4 :* Differentiate .

( 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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* SOLUTION 5 :* Differentiate .

( 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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* SOLUTION 6 :* Differentiate .

( The outer layer is ``the sine function'' and the inner layer is (5*x*) . Differentiate ``the sine function'' first, leaving (5*x*) unchanged. Then differentiate (5*x*) . ) Thus,

.

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* SOLUTION 7 :* Differentiate .

( The outer layer is ``the exponential function'' and the inner layer is . Recall that . Differentiate ``the exponential function'' first, leaving unchanged. Then differentiate . ) Thus,

.

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* SOLUTION 8 :* Differentiate .

( 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,

.

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* SOLUTION 9 :* Differentiate .

( 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 . )

.

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* SOLUTION 10 :* 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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* SOLUTION 11 :* Differentiate .

( 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,

.

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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,

.

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* SOLUTION 13 :* Differentiate .

( Since is a MULTIPLIED CONSTANT, we will first use the rule , where *c* is a constant. Hence, the constant just ``tags along'' during the differentiation process. 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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* SOLUTION 14 :* Differentiate .

( 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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* SOLUTION 15 :* Differentiate .

( 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 `` 7*x* plus the natural logarithm (base e) function'', and the fourth layer is (5*x*) . Differentiate them in that order. ) Thus,

.

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* SOLUTION 16 :* Differentiate .

( 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. )

.

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* SOLUTION 17 :* Differentiate .

( 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 differentiation process. 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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* SOLUTION 18 :* Differentiate .

( 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 (7*x*) . Differentiate them in that order. ) Thus,

.

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

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* SOLUTION 19 :* Assume that *h*(*x*) = *f*( *g*(*x*) ) , where both *f* and *g* are differentiable functions. If *g*(-1)=2, *g*'(-1)=3, and *f*'(2)=-4 , what is the value of *h*'(-1) ?

Recall that the chain rule states that . Thus,

so that

.

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* SOLUTION 20 :* Assume that , where *f* is a differentiable function. If and , determine an equation of the line tangent to the graph of *h* at *x*=0 .

The outer layer of this function is ``the third power'' and the inner layer is *f*(*x*) . The chain rule gives us that the derivative of *h* is

.

Thus, the slope of the line tangent to the graph of *h* at *x*=0 is

.

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

or .

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* SOLUTION 21 :* Determine a differentiable function *y* = *f*(*x*) which has the properties
and .

Begin with and assume that *f*(*x*) is not identically zero. Then

iff .

Note that

and

, where *C* is 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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Fri May 9 12:13:55 PDT 1997