1. ## Using Derivative rules...

I'm asked to find the slope of the tangent to the graph of the function at the point whose x-coordinate is given...

for the function: f(x)=x^-3(x^(-1)+1); x=1.

So simplifying...

f(x)=x^-3(x^(-1)+1)
f(x)=(1/x^3)(1/x+1)
f(x)=(1/3x^2)(2)
f(x)=2/3

Where do I go from here?

2. Originally Posted by Jeavus
I'm asked to find the slope of the tangent to the graph of the function at the point whose x-coordinate is given...

for the function: f(x)=x^-3(x^(-1)+1); x=1.

So simplifying...

f(x)=x^-3(x^(-1)+1)
f(x)=(1/x^3)(1/x+1)
f(x)=(1/3x^2)(2)
f(x)=2/3

Where do I go from here?
By f(x)=x^-3(x^(-1)+1) do you mean $f(x) = \frac{1}{x^3} \left( \frac{1}{x} + 1 \right) = \frac{1 + x}{x^4}$?

Whether you do or don't, the thing is that you've calculated f(1) when the question requires f'(1), that is, the value of the derivative of f(x) calculated at x = 1.

If my guess for your function is correct, then my simplification makes it clear that you should apply the quotient rule to get the derivative .....