3.9 Exponential and Log Functions

Find Derivative

100

Differentiate y = (lnx)²

y' = (2lnx/x)

Explanation: Use the chain rule to differentiate

200

Differentiate y = x^(3^x)

y = x^{(3^x)} 

lny = ln(x^{3^(x)})    

lny = 3^xln(x)  

(dy/dx)(1/y) = 3^x(1/x) + ln(x)(3^x)(ln3)

(dy/dx) = y((3^x/x) + ln(x)(3^x)(ln3)) 

                Answer:

(dy/dx) = x^{(3^x)} ((3^x/x) + ln(x)(3^x)(ln3))

Explanation:

Take natural log, drop the exponent, and solve for (dy/dx).

300

Differentiate y = (ln2x)^(ln9x)

y = (ln2x)^(ln9x)

lny = ln(9x)(ln(ln2x)) 

(1/y)(dy/dx) = ln9x(1/ln(2x))(2/2x) + (ln(ln2x))(9/9x)

(1/y)(dy/dx) = (2ln(9x)/2xln(2x)) + (9ln(ln(2x)/x)

        Final Answer:

(dy/dx) =  (ln2x)^(ln9x) ((2ln(9x)/2xln(2x)) + (9ln(ln(2x)/x)/ xln(2x))

Explanation:

Take the natural log of each side, drop the exponent, differentiate, and simplify.

400

Differentiate  g(x) = (17x)^(cos2x)

y = (17x)^(cos2x)

lny = ln((17x)^(cos2x)

lny = (cos2x)(ln(17x))

(1/y)(dy/dx) = (cos2x)(17/17x) + ln(17x)(-sin2x)(2)

       Final Answer:

(dy/dx) = 17x^(cos2x)( (cos2x/x) - 2(ln17x)(sin2x) )

Explanation:

Take the natural log of both sides, drop the exponent, differentiate, and solve for (dy/dx).

500

f(x) = log₃(5x-4)(2x)

f(x) = log₃(5x-4)(2x)

f'(x) = (1/(ln3(5x-4)(2x)) (2x(5) -(5x-4)(2))/(4x²)

f'(x) = (10x-10x + 8)/(ln3(5x-4)4x²)

f'(x) = 8/(ln3(5x-4)4x²)

Final Answer:

f'(x) = 4/(ln3(5x-4)x)

Explanation:

Use the change of base formula to differentiate and then simplify.

Find Derivative

100

Differentiate y = ln(lnx)

y' = (1/xlnx)

Explanation:

The derivative of ln(x) is (1/x). Use the chain rule to differentiate.

200

Differentiate  y = x^3x

y = x^3x

lny = ln(x^3x)

lny= 3xln(x)

(dy/dx)(1/y) = 3((x)(1/x) + lnx)

(dy/dx) = y(3((x)(1/x) + lnx)

         Final Answer:

(dy/dx) = x^3x(3 + 3lnx)

Explanation:

Take the natural log of both sides, drop the exponent, and solve for (dy/dx). Remember when you are solving that y = x^3x 

300

Differentiate  (d/dx)(e^x + e^-x) / (e^x + e^-x)

(d/dx)(e^x + e^-x) / (e^x + e^-x)

(dy/dx) = ((e^x + e^-x)(e^x + e^-x) - (e^x + e^-x)(e^x + e^-x) / ((e^x + e^-x)²

(dy/dx) = (e^2x + e⁰ + e⁰ + e^-2x - e^2x +e⁰  + e⁰ + e^2x) / ((e^x + e^-x)²)

       Final Answer:

(dy/dx) = 4/ ((e^x + e^-x)²)

Explanation: Use the quotient rule to differentiate. Remember that e⁰ = 1. Simplify by adding like terms together.

400

Differentiate f(x) = ln(3e^xx⁴)

f'(x) = (1/3(e^xx⁴) (3(e^x(4x³) + x⁴(e^x))

f'(x) = (3e^x(4x³ + x⁴))/(3e^xx⁴)

f'(x) = (4x³ + x⁴)/x⁴

f'(x) = x³(4+x)/(x⁴)

f'(x) = (4+x)/x

Final Answer:

f'(x) = (4/x) + 1

Explanation(using the chain rule):

1)Differentiate the outside function(which is an ln function so remember that (d/dx)lnx = (1/x))

2)Differentiate the inside function using the product rule

3)Factor out the common terms

4)Simplify

500

If f(x) = (4^1-5x)/(2-3x)

f'(x) = (2-3x)(4^1-5x)ln4(-5)-4^1-5x(-3)/(2-3x)²

f'(x) = -4^1-5x(5(2-3x)ln4)-3)/(2-3x)²

Explanation:

Use the quotient rule to differentiate. Remember the rule for differentiating a constant with an exponential function (a^ulna(du/dx))

Derivative of loga (u) & a^x

100

y = 3^2x

y' = ln3(3^2x)2

Explanation:

Use  derivative of a^u rules to differentiate

200

y = 4^{x^3 - sinx}

y' = ln4( 4^{x^3 - sinx})(3x² - cosx)

Explanation:

Use derivative of a^u rules to differentiate

300

y = 3^(3x+6) tanx

y' = 3^(3x+6)(sec²x) + tanx(ln3)(3^(3x+6)(3)

y' = 3^(3x+6)(sec²x + 3tanx(ln3))

Explanation:

1)Use derivative of a^u rules to differentiate

2) Simplify by factoring out  3^(3x+6)

400

y = logcosx

y' = ((1/(ln10 cosx))(-sinx)

Answer:

y'= (-sinx)/(ln10 cosx)

Explanation:

1) Use the formula to differentiate

2)Simplify  

500

Differentiate y = log₂(x²/x-1)

y = 2log₂x - log₂(x-1)

y' = 2(1/ln2)(1/x)(1) - (1/ln2)(1/x-1)(1)

y' = (2/xln2) - (1/ln2)(1/x-1)

y' = (1/ln2)((2/x) - (1/(x-1))

Answer:

y' = (1/ln2)((x-2)/x(x-1))

Explanation:

1)Use log properties to expand

2)Bring down the exponent

3)Differentiate

4)Simplify

Tangent Line

100

What is the tangent line equation for y = ln(√5-x²) at x = 2.

1) Differentiate the function

y' = (1/√5-x²)((1/2)(5-x²)^-1/2(-2x)

y' = (-2x)/(2(√5-x²)(√5-x²)

y' = -x/(5-x²)

2)Find the slope of the tangent line at x=2

y'(2) = 2/5-4 = 2

3)Plug in x=2 to the original function to find the y for the tangent equation

y = 3 + ln√5-4

y= 3 =ln1

y=3

4) Put all of the information together for the tangent equation

y-3=-2(x-2)

200

The tangent line to the graph of y = e^2-x at the point (1,e) intersects both coordinate axes. What is the area of the triangle formed by this tangent line and x- and y- axes?

1) Find the derivative of the function

 y' = e^2-x (-1)

y' = -e^2-x

2) Find the slope tangent line at x =1

-e^2-1=-e

3)Find the equation of the tangent line and solve for y

y - e = -e(x-1)

y=-ex +e +e

y=-ex + 2e

4) You know that the tangent line intersects both axes. Set the y = 0 to find the x coordinate of the x intercept

0=-ex + 2e

2e = ex

x=2

4) Set the x = 0 to find the y coordinate for the y intercept.

y=-e(0)+ 2e

y = 2e

5) The length of the y intercept to the origin is 2e (height) and the length of the x intercept to the origin is 2(base). Find the area of the triangle

A = 4e/2

A = 2e

300

Write the equation of the horizontal tangent(s) to the curve x² + 3y - lnx +y2 = x

(d/dx)(x² + 3y - lnx + y²) = (d/dx)(x)

2x + 3(dy/dx) - (1/x) + 2y(dy/dx) = 1

3(dy/dx) + 2y(dy/dx) = 1 + (1/x) -2x

(dy/dx)(3+2y) = (x+1-2x²)/x

(dy/dx)=(-2x² +x +1)/(x(3+2y))

(-2x² +x +1)/(x(3+2y)) = 0

(-2x² +x +1) = 0

(-2x-1)(x-1) = 0

x=-(1/2) x=1

(x=-(1/2) will not work because ln(-1/2) does not exist)

1² + 3y - ln(1) + y² = 1

3y + y² = 0

y(3 + y) = 0

y = 0 y = -3

The horizontal tangent cannot be y = 0, so y = -3 is the horizontal tangent.

Explanation:

1)Use implicit differentiation and solve for (dy/dx). This is the derivative.

2)Set this equation equal to 0 and solve for x to find the point where the horizontal tangent is at

3) Your answer is x=-(1/2) x=1, but x=-(1/2) will not work because ln(-1/2) does not exist

4) Plug in  x=1 to the original equation. You get      y = -3. This is the equation of the horizontal tangent.

400

The slope of the line tangent to the graph of the graph of e^xy=2 at the point where x =1 is

e^xy=2

ln(e^xy) = ln2

xy = ln2

y = (ln2/x)

y ' = (x(0) - ln2)/x²

y' = -(ln2)/x²

y'(1) = -(ln2)/1

y'(1) = ln2

The slope of the tangent line is -ln2

Explanation:

1)Differentiate  e^xy=2 by taking the ln of both sides

2)lne = 1, so you are left with xy = ln2

3)Solve for y

4)Differentiate

5)Plus in x =1 to find the derivative at this point, which is the slope of the tan line

500

Find where the tangent line is horizontal to the function f(x) = (2x-4)e^x. What is the equation of the tangent line.

f'(x) = (2x-4)e^x + e^x(2)

f'(x) = 2e^x((x-2) + 1) 

f'(x) = 2e^x(x-1)

2e^x(x-1) = 0

(x-1) = 0

x=1 

When x=1, the tangent line is horizontal to the function.

y = (2(1)-4)e⁽¹⁾

y = -2e

The equation of the tangent line is y = -2e

Key Terms

100

What is the inverse of the natural log function?

f(x) = ln x is f^-1(x) = e^x.

200

How do you use natural log differentiation?

1) Take natural log of both sides  of equation y = f(x) and use laws of logs to simplify

2)Differentiate with respect to x

3)Solve for y'

4)Replace y with f(x)

300

The log function with base a, where a>0  and a does not equal 1 is

Y = log_ax if and only if x = a

400

How do you apply the change of base formula when differentiating?

(d/dx)(log_ax) = (1/lna(x))

(d/dx)(log_au) = (1/lna(u))(du/dx)

500

How do you differentiate exponential functions? 

(d/dx)e^u = e^u(du/dx)

(d/dx)a^u = a^ulna(du/dx)