1. What is the equation of the tangent line to y = 2*x(x - 4)^6 at point (5, 10)?

2. If f(x)=8/(x^2+4) , find the slope of the curve at the point (2, 1).

3. The demand function for an item is given by x=200[1-(p/(p+2))] . At what rate is the demand changing when the price p is $8.00 per item?

Which two do you need help with? You did say "a couple"

For question 1, take the derivative of y(x), dy/dx, and calculate its value m at x = 5. Then the equation for your tangent line will be
(y - 10) = m (x - 5).

For question 2, calculate the value of the derivative df/dx at x=2.

For question 3, calculate dx/dp wnen p = 8.

We will be happy to provide further assistance if needed. Please show your work

ok annie, for 1. we are given y(x)and are asked for the tan. line at some point. You do know, or should, that the derivative of y will give us the slope at a point for y. However, I think that the exercise is also expecting you use a theorem. You could expand the poly (x-4)^6, or make use of the theorem for the product of 2 functions. Expanding the poly is tedious unless you can see by observation alone that (x-4)^6 = X^6 - 4x^5 + 16x^4 - 64x^3 + 256x^2 - 1024x +1, as I or an experienced math type can? Thus I think they want you to see y = f(x)g(x) where f = 2x and g = (x-4)^6 and use the product rule. You're text should show that. Substitute 5, the x value, to find the slope. You now have a point and the slope and should be able to handle the equation for the line.
Question 2. is similar, except that the questioner here expects you to know how to handle a function in the denom. Here is a case where it is easier to think of 8/(x^2+4) as being the same as 8(x^2 + 4)^-1 . This simplifies how you might think of the derivative. Once again, evaluate the equation for x = 2 to get the slope. As before, you now have the slope and a point and should be able to go from there.
For 3. you might want to think of x = 200(1 - p(p+2)^(-1)). Use the product rule. Incidentally, in this example the questioner wants you to think of x as the dependent variable. They don't want you to always be thinking of y(x), f(x), etc, but beaware that we can have x(p) too. Find the derivative and evaluate it at p = 8.

hmm.., After review I see I got the constant term in 1. wrong. It should be 4096. As I said, I did that one by 'eyeball' alone only because I know the powers of 2 well. But I think I edited something out and wrote over the constant term. My point however is that I doubt that's what they wanted you to do. Be sure to study the theorems and be alert to where you can apply them.

For question 1, you are given the equation y = 2*x(x - 4)^6 and asked to find the equation of the tangent line at the point (5, 10). To do this, we need to find the derivative of y with respect to x, which will give us the slope of the tangent line at any point on the curve.

Taking the derivative of y with respect to x, we can use the product rule. Let's define f(x) = 2*x and g(x) = (x - 4)^6. Applying the product rule, we have:

y' = f'(x)*g(x) + g'(x)*f(x) = 2*(x - 4)^6 + 6*2*x*(x - 4)^5

Simplifying this expression, we get:

y' = 2*(x - 4)^6 + 12*x*(x - 4)^5

Substituting x = 5 into the derivative, we can find the slope of the tangent line at that point:

m = y'(5) = 2*(5 - 4)^6 + 12*5*(5 - 4)^5 = 2*(1)^6 + 12*5*(1)^5 = 2 + 12*5 = 62

The slope of the tangent line is 62. Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line. We have the point (5, 10) and the slope 62. Using the equation:

(y - y1) = m(x - x1)

where (x1, y1) is the point (5, 10), we can substitute the values and simplify:

(y - 10) = 62(x - 5)

This is the equation of the tangent line to y = 2*x(x - 4)^6 at the point (5, 10).

For question 2, you are given the function f(x) = 8/(x^2 + 4) and asked to find the slope of the curve at the point (2, 1). To do this, we need to find the derivative of f(x) with respect to x.

To find the derivative, we can use the quotient rule. Let's define f(x) = 8 and g(x) = x^2 + 4. Applying the quotient rule, we have:

f'(x) = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2

Simplifying this expression, we get:

f'(x) = (0 - 8*(2*x))/(x^2 + 4)^2 = -16x/(x^2 + 4)^2

Substituting x = 2 into the derivative, we can find the slope of the curve at that point:

m = f'(2) = -16*2/(2^2 + 4)^2 = -32/36 = -8/9

The slope of the curve at the point (2, 1) is -8/9.

For question 3, you are given the demand function x = 200[1 - (p/(p + 2))] and asked to find the rate at which the demand is changing when the price p is $8.00 per item. To do this, we need to find the derivative of x with respect to p.

To find the derivative, we can use the chain rule. Let's define f(p) = 200 and g(p) = 1 - (p/(p + 2)). Applying the chain rule, we have:

x' = f'(g(p))*g'(p)

We can calculate the derivatives:

f'(p) = 0 (since f(p) is a constant function)
g'(p) = -[(p + 2) - p]/(p + 2)^2 = -2/(p + 2)^2

Substituting these values into the derivative expression, we get:

x' = 0*(-2/(p + 2)^2) = 0

Therefore, the rate at which the demand is changing when the price is $8.00 per item is 0.