• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Help me find derivatives

Omegachi

Diamond Member
can any one help me find the 1st and 2nd derivative of

117e^(-10/x)

THen can you solve for X for both of the deriv...
 
So you have f(x)=117e^(-10/x). Looks like a function of the form Ae^u, where u=(-10/x). Since the deriv. of e^x is itself, the deriv of e^u is e^u (du/dx) via Chain Rule.

I get f'(x)= 117e^(-10/x)(10/(x^2)), where the second parentheses is from Quotient Rule on the 'u' part.

Cleaning up yields f'(x) = (1170/(x^2)) e^(-10/x) Hopefully I didn't goof 😀

A characteristic of exponential fcns. is that when you differentiate, the e raised to the whatever part appears in the derivative, and some extra junk is created also, in the form of additional terms out in front.

To find f ' ', I think it's more involved. If you look at the first deriv. compared to the original function, the term out in front now has x in it, where before it was just a constant. So, you'll have to use Product Rule to differentiate, with the two parts (1170/(x^2)) and e^(-10/x) as your f(x) and g(x). And that gets hairy.

As for solving, well what are the derivatives set equal to? Zero? Some other constant? It just becomes an algebraic exercise.
 
Back
Top