### 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) .

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,

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,

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.)

.

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.)

.

SOLUTION 6 : Differentiate .

( 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,

.

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,

.

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,

.

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 . )

.

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,

.

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,

.

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

SOLUTION 12 : Differentiate .

( 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,

.

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 . )

.

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,

.

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

.

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. )

.

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. )

.

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

.

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

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

.

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 .

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,

.