121616949-math.37 - 2.1 The slope of a function 23 which is...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
2.1 The slope of a function 23 which is the slope of a line, then we figure out what happens when Δ x gets very close to 0. We should note that in the particular case of a circle, there’s a simple way to find the derivative. Since the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, its slope is the negative reciprocal of the slope of the radius. The radius joining (0 , 0) to (7 , 24) has slope 24/7. Hence, the tangent line has slope - 7 / 24. In general, a radius to the point ( x, 625 - x 2 ) has slope 625 - x 2 /x , so the slope of the tangent line is - x/ 625 - x 2 , as before. It is NOT always true that a tangent line is perpendicular to a line from the origin—don’t use this shortcut in any other circumstance. As above, and as you might expect, for different values of x we generally get different values of the derivative f ( x ). Could it be that the derivative always has the same value?
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern