(a) The function f ( x , y ) = x y x 2 + y 2 if ( x , y ) ≠ ( 0 , 0 ) 0 if ( x , y ) = ( 0 , 0 ) was graphed in Figure 4. Show that f x ( 0 , 0 ) and f y ( 0 , 0 ) both exist but f is not differentiable at ( 0 , 0 ) . [Hint: Use the result of Exercise 53.] (b) Explain why f x and f y are not continuous at ( 0 , 0 ) .
(a) The function f ( x , y ) = x y x 2 + y 2 if ( x , y ) ≠ ( 0 , 0 ) 0 if ( x , y ) = ( 0 , 0 ) was graphed in Figure 4. Show that f x ( 0 , 0 ) and f y ( 0 , 0 ) both exist but f is not differentiable at ( 0 , 0 ) . [Hint: Use the result of Exercise 53.] (b) Explain why f x and f y are not continuous at ( 0 , 0 ) .
Solution Summary: The author explains that the function f isn't differentiable at (0,0).
f
(
x
,
y
)
=
x
y
x
2
+
y
2
if
(
x
,
y
)
≠
(
0
,
0
)
0
if
(
x
,
y
)
=
(
0
,
0
)
was graphed in Figure 4. Show that
f
x
(
0
,
0
)
and
f
y
(
0
,
0
)
both exist but
f
is not differentiable at
(
0
,
0
)
. [Hint: Use the result of Exercise 53.]
(b) Explain why
f
x
and
f
y
are not continuous at
(
0
,
0
)
.
Consider the function f(x) = x²-1.
(a) Find the instantaneous rate of change of f(x) at x=1 using the definition of the derivative.
Show all your steps clearly.
(b) Sketch the graph of f(x) around x = 1. Draw the secant line passing through the points on the
graph where x 1 and x->
1+h (for a small positive value of h, illustrate conceptually). Then,
draw the tangent line to the graph at x=1. Explain how the slope of the tangent line relates to the
value you found in part (a).
(c) In a few sentences, explain what the instantaneous rate of change of f(x) at x = 1 represents in
the context of the graph of f(x). How does the rate of change of this function vary at different
points?
1. The graph of ƒ is given. Use the graph to evaluate each of the following values. If a value does not exist,
state that fact.
и
(a) f'(-5)
(b) f'(-3)
(c) f'(0)
(d) f'(5)
2. Find an equation of the tangent line to the graph of y = g(x) at x = 5 if g(5) = −3 and g'(5)
=
4.
-
3. If an equation of the tangent line to the graph of y = f(x) at the point where x 2 is y = 4x — 5, find ƒ(2)
and f'(2).
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.