f: R2-->R is defined by f(x) = absolute value of x. Find Df(1,0).
So Df(x)=[x/sqrt.(x^2), 0]
So Df(1,0)= 1/sqrt.1^2=1
Did I do this right?
Do you mean f(x,y)= |x|? Or $\displaystyle f(x,y)= \sqrt{x^2+ y^2}$?
If you meant f(x, y)= |x|, for x positive |x|= x so the derivative with respect to x is 1 and the derivative with respect to y is 0. $\displaystyle \nabla f(1, 0)= <1, 0>$.So Df(x)=[x/sqrt.(x^2), 0]
So Df(1,0)= 1/sqrt.1^2=1
Did I do this right?
If you meant $\displaystyle f(x,y)= \sqrt{x^2+ y^2}$, then $\displaystyle \nabla f= \left<\frac{x}{\sqrt{x^2+ y^2}}, \frac{y}{\sqrt{x^2+ y^2}}\right>$ so $\displaystyle \nabla f(1, 0)= <1, 0>$.