Recent content by sl2382

I think the requirements are: 1)the derivative exists at a point 2)limits approaching from both sides of that point are the same ? So, the derivative does not exist at 0, BUT isn't it defined at 0 ? Does that mean the derivative actually exists and the function can be differentiate?

Homework Statement B(x)= xsin(1/x) when x is not equal to 0 = 0 when x is equal to 0 Determine if the function is differentiable at 0 Homework Equations The Attempt at a Solution I get B'(x)= sin(1/x)+cos(1/x)*(-1/x) but really do not know what should be done next...

Thank you so much for all of your helps! Thanks!

Just some general questions as I'm confused with when to use chain rule when not to. For instance, to find the derivative of e^sqrt(x), the right answer is to use chain rule to get e^sqrtx*the derivative of sqrt(x). BUT, isn't there a formula that: d/dx K^x = In(K)*K^x? K for constant and x...

Oh I got it. Thank you so much!

Does one-to-one means that g(x) can't be 0? Since if g=0 then it can't pass the horizontal line test.

Is it -g'(x)/g^2(x) ?

Hi, thank you so much. I calculated once again: g^2(x)'/g^2(x) Is this one right? It is the reciprocal, the 1/g(x). The question asks for if 1/g(x) is differentiable at every point in its domain.

Homework Statement If g(x) is one-to-one and differentiable at every point in R, then its reciprocal 1/g(x) is also differentiable at every point in its domain. True of False? Homework Equations The Attempt at a Solution I have tried several times, and the statement seems to be...