Name______

Directions: For each category see if you can devise the rule(s) that was used to find the derivative(s) and then use it to find the derivatives of the functions that follow.

Category 1: ex

1. Function: fx=ex 3. Function: fx=e5x

Derivative: f'x=ex Derivative: fx=5e5x

2. Function: fx=4ex2 4. Function: fx=ex3-2x

Derivative: f'x=8xex2 Derivative: f'x=(3x2-2)ex3-2x

What is the formula for the derivative of eu? What role does the chain rule play?

The derivative of ex is ex, and the chain rule is used on the u term of eu.

Find the derivative of the following:

1. fx=5e7x 2. fx=e9x2+6x

f'(x)=7*5e7x=35e7x fx=e9x2+6x

Combining Rules

1.  If fx=xe3x what rule that we have previously learned should we use to find its derivative? Use that rule to find the derivative of f(x).

The product rule.

f'(x)=xe3x=e3x+3xe3x

2.  If fx=e2xsin⁡(5x) what rule that we have previously learned should we use to find its derivative? Use that rule to find f’(x).

The quotient rule, and the sine rule.

f'x=2e2xsin5x-10e2xcos5xsin2(5x)

3.  If fx=arctan⁡(ex4)what rule that we have previously learned should we use to find its derivative? Use that rule to find f’(x).

The chain rule and the arctan rule.

f'x=4x3ex41+e2x4

Category 2: ln(u)

1.  Function: fx=ln⁡(3x) 3. Function: fx=ln(x2)

Derivative: f'x=33x=1x Derivative: f'x=2xx2=2x

2.  Function: fx=ln⁡(x3+4) 4. Function: fx=ln⁡(sin2x)

Derivative: f'x=3x2x3+4 Derivative: f'x=2cos⁡(2x)sin⁡(2x)=2cot⁡(2x)

What is the formula for the derivative of ln(u)?

The derivative of ln(u) is the derivative of the term divided by term itself.

Find the derivatives of the following:

1.  fx=2ln(3x) 2. fx=ln(5x2-4x+1)

f'(x)=2*33x=63x=2x f'x=10x-45x2-4x+1

Category 3: Logarithmic expansions

1.  Function: fx=ln⁡(xx+1) 3. Function: fx=ln⁡(x2tan5x)

Expansion: fx=lnx+12ln⁡(x+1) Expansion: fx=2lnx+ln⁡(tan5x)

Derivative: f'x=1x+12(x+1) Derivative: f'x=2x+5sec2(x)tan⁡(5x)

2.  Function: fx=lnx2x-1 4. Function: fx=lnxcsc⁡(x)

Expansion: fx=lnx2-ln⁡(x-1) Expansion: fx=12lnx-ln⁡(secx)

Derivative: f'x=2x-1x-1 Derivative: f'x=12x-secxtan⁡(x)sec⁡(x)=12x-tan⁡(x)

What rules are governing the expansion of the function so that we can use the formula for

the derivative of ln(u) to find the derivative?

The rules of ln are governing the expansion of the terms (i.e. ln(a/b) = ln(a) – ln(b) and ln(a*b) = ln(a) + ln(b)). Once the expansion is performed, the ln(u) rule is applied to all terms in the expanded function.

Find the derivatives of the following:

1.  fx=ln⁡[(x2-4)(x+5)] 2. fx=lnx3-73x

fx=lnx2-4+ln(x+5) fx=lnx3-7-ln⁡(3x)

f'x=2xx2-4+1x+5 f'(x)=3x2x3-7-13x-233x=3x2x3-7-139x

3. fx=ln⁡(x3ex2x-9) 4. fx=ln⁡[x4cosx+sinx

fx=lnx3ex-ln⁡(2x-9) fx=ln⁡(x4)+ln⁡cosx+sinx

f'(x)=3x2ex+x3exx3ex-22x-9=3ex+xexxex-22x-9 f'(x)=4x3x4+cosx-sin⁡(x)cosx+sin⁡(x)

Category 4: L’Hospital’s Rule

1. limx→0ex-1x = (by L’Hospital’s Rule) limx→0ex1=1

2. limx→0cosx-12sin⁡(x)= (by L’Hospital’s Rule) limx→0sin⁡(x)2cos⁡(x)=02=0

3. limx→∞2x2-3x+45x2-8= (by L’Hospital’s Rule) limx→∞4x-310x= (by L’Hospital’s Rule) limx→∞410= 25

What rule or process are we using to find the indicated limits? Why are we learning about this rule for finding limits after we have studied derivatives? What does L’Hospital’s Rule seem to allow us to do? Under what circumstances can we use it? (Hint: What do the functions look like if you try to plug and play zero or infinity into them?)

This rule is taking the derivative of the numerator and denominator until the x value for which we seek to find the limit can be substituted into the equation that we have created. At that point we substitute in the x value and the solution is the limit at the specified x value.

Find these limits

1. limx→12ex-2ln⁡(x)

limx→12ex-2ln⁡(x)=by L'Hospital's Rulelimx→1ex+ex1x=limx→12xex=2e

2. limx→0ex-e-x4sin⁡(x)

limx→0ex-e-x4sin⁡(x)=by L'Hospital's Rulelimx→0ex+e-x4cosx=24=12

Part II: Use your Calculator to determine the following:

1.  The slope of the tangent to the curve at the point (4, -2.234).

This is equal to the slope of f(x) at x = 4, which is f’(4).

f'x=x3-6x52+5x2x3+10 f'4=43-6*452+542*43+10=-0.44565

2.  Evaluate the derivative of at x = 1

-12.723, the line tangent to f(x) has a negative slope at x=1.

3.  Plot f(x) =ln(x) and f’(x) = 1/x in your calculator. Remembering that the derivative of f(x) is the instantaneous rate of change of f(x) why does it make sense that 1/x is the derivative of ln(x)?

f(x) =ln(x) is red

f’(x) = 1/x is blue

As f(x) increases the slope decreases towards zero, and 1/x, as x increases, approaches zero as well.

4.  Graph y = tan(x) and its derivative together on (-π/2, π/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for you answers?

f(x) is red

f’(x) is blue

The tangent graph does appear to have a smallest slope, at 0. The largest slope occurs as the tangent function approaches π/2. The slope is never negative because this function repeats itself every π/2 x values. Also, the tangent function is undefined as tangent nears π/2.