Practice problems for these solutions can be found at: Differentiation Practice Problems for AP Calculus

**Part A The use of a calculator is not allowed.**

- Applying the power rule, =30
*x*^{4}– 1. - Rewrite
*f (*= as*x*)*f (*=*x*)*x*+^{–1}*x*^{–2/3}. Differentiate,*f'*(*x*)= –*x*^{–2}–*x*^{–5/3}= – - Rewrite
- Applying the quotient rule,
- Applying the product rule,
*u*=(3*x*– 2)5 and*v*=(*x*2 – 1), and then the chain rule, - Let
*u*=2*x*– 1, - Using the product rule,
=(3[sec(3

*x*)])+[sec(3*x*) tan(3*x*)](3)[3*x*]=3 sec(3

*x*)+9*x*sec(3*x*) tan(3*x*)=3 sec(3

*x*)[1+3*x*tan(3*x*)] - Using the chain rule, let
*u*= sin(*x*^{2}– 4). - Using the chain rule, let
*u*=2*x*. - Since 3e 5 is a constant, thus its derivative is 0.
- Using implicit differentiation, differentiate each term with respect to
.*x* - Since f' (4) is equivalent to the slope of the tangent to f (
*x*) at*x*=4, there are several ways you can find its approximate value. - You can use the difference quotient to approximate f' (a).
- Enter
*y*1=*x*5 +3*x*– 8. The graph of*y*1 is strictly increasing. Thus f (*x*) has an inverse. Note that f (0)= – 8. Thus the point (0, –8) is on the graph of f (*x*) which implies that the point (–8, 0) is on the graph of f –1(*x*). - =(2)(sin
*x*)+(cos*x*)(2*x*)= 2 sin*x*+2*x*cos*x* - Enter
*y*1=(*x*– 1)2/3 +2 in your calculator. The graph of*y*1 forms a cusp at*x*=1. Therefore, f is not differentiable at*x*=1. - Differentiate with respect to
*x*:

*f'* (*x* )=[5(3*x* – 2)^{4}(3)][*x*^{2} – 1]+[2*x* ] × [(3*x* – 2)^{5}]

=15(*x*^{2} – 1)(3*x* – 2)^{4} +2*x* (3*x* – 2)^{5}

=(3*x* – 2)^{4}[15(*x*^{2} – 1)+2*x* (3*x* – 2)]

=(3*x* – 2)^{4}[15*x*^{2} – 15+6*x*^{2} – 4*x* ]

=(3*x* – 2)^{4}(21*x*^{2} – 4*x* – 15)

=10[–csc^{2}(2*x* – 1)](2)

= – 20 csc^{2}(2*x* – 1).

=10(–sin[sin(*x*^{2} – 4)])[cos(*x*^{2} – 4)](2*x* )

= – 20*x* cos(*x*^{2} – 4) sin[sin(*x*^{2} – 4)]

=0+(4)(e *x* )+(e *x* )(4*x* )

=4e *x* +4*x*e *x* =4e *x* (1+*x* )

**Part B Calculators are allowed.**

Method 1: Using the slope of the line segment joining the points at *x* =4 and *x* =5.

- f (5)=1 and f (4)=4

Method 2: Using the slope of the line segment joining the points at *x* =3 and *x* =4.

- f (3)=5 and f (4)=4

Method 3: Using the slope of the line segment joining the points at *x* =3 and *x* =5.

- f (3)=5 and f (5)=1

Note that –2 is the average of the results from methods 1 and 2. Thus f' (4) ≈ –3, –1, or–2 depending on which line segment you use.

Or, you can use the symmetric difference quotient to approximate f' (a).

Thus, f' (2) ≈ 4 or 5 depending on your method.

Thus the slope of the tangent to*y*= ln *x* . At *x* =e,*y*= ln *x* = ln *e* = 1, which means the point (e, 1) is on the curve of*y*= ln *x*. Therefore, an equation of the tangent is *y* – 1= at *x* =e is (*x* – e) or*y*= . [See Figure 6.11-1.]

- The expression is the derivative of sin
*x*at*x*=π/2 which is the slope of the tangent to sin*x*at*x*=π/2. The tangent to sin*x*at*x*=π/2 is parallel to the*x*-axis. - Using the chain rule, let
*u*=(π –*x*). - Since the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator, the limit is 0.
- You can use the difference quotient to approximate
*f'*(*a*). - (See Figure 6.12–1.) Checking the three conditions of continuity:
- f (3) = 3

Therefore the slope is 0, i.e., = 0. An alternate method is to expand sin(π/2+*h*) as sin(π/2) cos *h* + cos(π/2) sin *h*.

Then, *f'* (*x* )=2 cos(π – *x* )[–sin(π – *x* )](–1)

=2 cos(π – *x* ) sin(π – *x* )

*f'*(0)=2 cos π sin π =0.

Or, you can use the symmetric difference quotient to approximate *f'* (*a*).

Thus, *f'* (2)=1.7, 2.05 or 1.25 depending on your method.

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