Question

Slope of hyperbola= 1/x. In finding the tangent for the Single Variable Calculus Video on Rate of change, shouldn't the difference quotient be (f(x0)-f(x0+changex))/changex? The reason his difference quotient I feel did not work was because his example involved a positive slope, where as the tangent line for the example results in a negative slope.

Answer 1

\(f(x)=1/x,\ f^'(x)=-1/x^2 \)
Or, if:
\(f^'(x)=1/x,\ f(x)=ln(x) +C \)

Answer 2

Thanks for response this helped!