Derivative at a Point: Tangent Line and Demand Curve Equations

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Homework Statement


a)Find the equation of the tangent line to f(x) = (x - 6)^8 at the point where x = 7.




b)The demand curve for a product is given by
q = f(p) = 5,000e-0.30p,
where q is the quantity sold and p is the price of the product, in dollars. Find f(6) and f'(6).



Homework Equations


a)sustitute x for 8

b) to find F(6) i sutitute p for 6
to find F'(6) i sustitute p for 6?




The Attempt at a Solution



a)the solution is 1 and this is incorrect

b)in F(6)=826 this is correct
in F'(6)=247.94 this is incorrect
 
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Substitute x for 8? What does that mean?

Show us more of your work for finding the derivatives of these functions. Maybe we can point out what you're doing wrong.
 
Keep in mind that you're finding an equation for part (a), and not a number.

As hage567 said showing some work would be beneficial to all.
 
What is the derivative of (x- 6)8?
 
Part A: If x=7 is the point of tangency, the slope of the tangent line to the function is the derivative of that function evaluated at the point x. So first find the derivative of your function, then evaluate it at the point of tangency. Given the slope, you can find the equation of the line given a set of points on that line.
 
It's clear that, after having posted this same problem repeatedly and being told, repeatedly, that he must show some effort himself, lolo105 is still expecting someone to do the problem for him!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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