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**Differentiation ---Help, please!**

**1. Homework Statement**

1) Use Newtons Method to approximate the real zero of the function f(x)=x^3 + 7x + 3 = 0

2) The curve y = x^3-2x^2+x-3 intersects the curve y = cos(2x). If x = 1.8 is used as the first estimate then, using Newton's Method, what is the x-value or the intersection point, accurate to 4 decimal places?

3) A curve is given by y = a^3, where a = m^2(2m-1). Approximate the slope of the curve when m = 0.6543

4) If y = -4 / [tex]\sqrt[3]{x+5}[/tex], then dy/dx

5) Find an equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

**2. Homework Equations**

1) Newtons Method:

x = f(x) / f'(x)

Or Slope = Rise/Run

2) Newtons Method

x = f(x) / f'(x)

Or Slope = Rise/Run

3) y = mx+b

derivatives

4) Quotient Rule

(f'g-g'f) / (g^2)

5) Quotient Rule

(f'g-g'f) / (g^2)

y=mx+b

**3. The Attempt at a Solution**

1) x^3 + 7x = 0

Let the "guess" be "n"

Let the next "guess" to be "n2" (2 was suppose to be a subscript but I can't find it in the Latex Reference)

Slope: derivative of x^3 + 7x = 0

3x^2 + 7 = 0

Rise: Plug in the "n" to the equation

n^3 + 7n = 0

Run: "n" subtract the "n2"

n - n2

Final formula:

3x^2 + 7 = n^3 + 7n / n - n2

*The answer should be this: (I was close but my rise is really off)

The first "n" is suppose to be a lower subscript. I can't find it here in the Latex Reference.

xn +1 = 2x[tex]^{3}_{n}[/tex] -3 / 3x[tex]^{2}_{n}[/tex] + 7

2) x^3 - 2x^2 + x - 3 - cos(2x) = 0

Slope = Rise/Run

Slope: Derivative of x^3 - 2x^2 + x - 3 - cos(2x)

3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2)

Rise: Plug in 1.8 to the equation

(1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8))

Run: 1.8 - x

Final Equation:

3(1.8)^2 - 4(1.8) + 1 + sin2(1.8)(2) = (1.8)^3-2(1.8)^2 + 1.8 -3 - cos(2(1.8)) / 1.8 - x

Simplify: 3.645639188 = -2.846026728 / 1.8 - x

-2.846026728 / 3.645639188 = 1.8 - x

1.8 + 0.780666045 = x

x = 2.680666046

The answer in the book is 2.0923.. So, do you think I should use my "x" to be the second trial and do the whole equation again from the start to get the 2.0923?

3)

a = m^2(2m-1)

(0.6543)^2(2(0.6543)-1))

a= 0.13211428

y = a^3

Therefore...

y = 0.13211428^3

y =0.002305946

I'm stuck. Please help?

4) ((0)([tex]\sqrt[3]{x+5}[/tex])) - ((-4)(1/3(x+5)^-2/3)) / ([tex]\sqrt[3]{x+5}[/tex])^2)

Simplify: ((4)(1/3(x+5)^-2/3) / ([tex]\sqrt[3]{x+5}[/tex])

Then I don't know how to simplify it much longer.. My simplifying skills has gone bad, I guess. Help, please?

5) Get y-value, plug in the x-value which is 1

1-1 / 1+1 = 0

(1,0)

Do quotient rule on the equation...

(1)(x+1) - (x-1)(1) / (x+1)^2

Simplify...

(x+1) - (x-1) / (x^2 + 2x+2)

Plug in the x-value which is 1 to get the slope.

(1+1) - (1-1) / (1^2 +2(1) + 2)

2 / 1+2+2 = 2 / 5

Slope = 2/5

y=mx+b

0 = 2/5(1) + b

0 - 2/5 = b

-2/5 = b

Final equation...

y = 2/5x -2/5

Although it seems like I did everything right, my final equation seems to be incorrect from the answer in the book which is x-2y=1

THAT'S ALL FOR NOW. I'll continue more questions later on if I find any that bugs my mind. But for now, I need to go to church so in the meanwhile, PLEASE HELP? =/