![To be differentiable, a function must be continuous and smooth. Derivatives will fail to exist at: cornercusp vertical tangent discontinuity. - ppt download To be differentiable, a function must be continuous and smooth. Derivatives will fail to exist at: cornercusp vertical tangent discontinuity. - ppt download](https://images.slideplayer.com/32/9955957/slides/slide_6.jpg)
To be differentiable, a function must be continuous and smooth. Derivatives will fail to exist at: cornercusp vertical tangent discontinuity. - ppt download
![SOLVED: Decide if the following function f(x) is differentiable at x=0. Try zooming in on a graphing calculator (like desmos or geogebra), or try calculating the derivative f^'(0) from the definition. f(x)={ SOLVED: Decide if the following function f(x) is differentiable at x=0. Try zooming in on a graphing calculator (like desmos or geogebra), or try calculating the derivative f^'(0) from the definition. f(x)={](https://cdn.numerade.com/ask_images/e54f11aa210041c89ffbce888c9de3b5.png)
SOLVED: Decide if the following function f(x) is differentiable at x=0. Try zooming in on a graphing calculator (like desmos or geogebra), or try calculating the derivative f^'(0) from the definition. f(x)={
![SOLVED: Question 1 - Differentiability (Show Working) Let f : â„ â†' â„ be given by: f(x,y) = √(x^2 + y^2) if (x,y) ≠(0,0) f(x,y) = 0 if (x,y) = (0,0) SOLVED: Question 1 - Differentiability (Show Working) Let f : â„ â†' â„ be given by: f(x,y) = √(x^2 + y^2) if (x,y) ≠(0,0) f(x,y) = 0 if (x,y) = (0,0)](https://cdn.numerade.com/ask_images/ed0878f322fd4e929ffebdf1a34d3647.jpg)