Show that the function f(x, y) = xy – 6y2 is differentiable by finding values of ɛ, and ɛ, that satisfy the following definition. Definition: If z = f(x, y), then f is differentiable at (a, b) if Az can be expressed in the form Az = f,(a, b) Ax + f,(a, b) Ay + 8,Ax + 8,Ay, where s, and 8, → 0 as (Ax, Ay) – (0, 0). To find an expression for Az, we choose an arbitrary point (a, b), another point arbitrarily close to the first (a + Ax, b + Ay), and compute the difference between their respective f-values. Az f(a + Дх, b + Ду) - f(a, b) %3D (а + Дх)(b + Ду) - 6(Ь + Дy)? - = ab + aly + bAx + AXAY – 6b2 - 12bay - 6(Ay)2 - ab + Ay + bAx + AxAy - AyA but f,(a, b) = and f(a, b) = , So Az = f,(a, b)Ax + f,(a, b)Ay + AxAy – 6AyAy. Let ɛ, = Ay and ɛ, = -6Ay, therefore f is differentiable by the given definition.
Show that the function f(x, y) = xy – 6y2 is differentiable by finding values of ɛ, and ɛ, that satisfy the following definition. Definition: If z = f(x, y), then f is differentiable at (a, b) if Az can be expressed in the form Az = f,(a, b) Ax + f,(a, b) Ay + 8,Ax + 8,Ay, where s, and 8, → 0 as (Ax, Ay) – (0, 0). To find an expression for Az, we choose an arbitrary point (a, b), another point arbitrarily close to the first (a + Ax, b + Ay), and compute the difference between their respective f-values. Az f(a + Дх, b + Ду) - f(a, b) %3D (а + Дх)(b + Ду) - 6(Ь + Дy)? - = ab + aly + bAx + AXAY – 6b2 - 12bay - 6(Ay)2 - ab + Ay + bAx + AxAy - AyA but f,(a, b) = and f(a, b) = , So Az = f,(a, b)Ax + f,(a, b)Ay + AxAy – 6AyAy. Let ɛ, = Ay and ɛ, = -6Ay, therefore f is differentiable by the given definition.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
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Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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![Show that the function f(x, y) = xy – 6y2 is differentiable by finding values of ɛ, and ɛ, that satisfy the following definition.
Definition: If z = f(x, y), then fis differentiable at (a, b) if Az can be expressed in the form Az = f,(a, b) Ax + f,(a, b) Ay + 8,Ax + 8,Ay, where &, and s, →
(Дх, Ду) — (0, 0).
To find an expression for Az, we choose an arbitrary point (a, b), another point arbitrarily close to the first (a + Ax, b + Ay), and compute the difference between their respective
f-values.
Дz 3 f(a + Дх, b + Ду) - f(a, ь) %3D (а + Дх) (b + Ду) - 6(b + ду)? -
— ab + aдy + ьдх + ДхДу - бь? - 12bДу - 6(Ду)2- ab +
- (L
+ bAx + AxAy -
but f,(a, b) =
and f,(a, b) =
, so
Az = f,(a, b)Ax + f,(a, b)Ay + AXAY – 6AyAy.
Let ɛ, = Ay and ɛ, = -6Ay, therefore f is differentiable by the given definition.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F632065a4-87e4-4db9-9a16-90da95c730a4%2F80d13b7a-4635-4736-9d2c-b05174a1da27%2Fl8xgb6_processed.png&w=3840&q=75)
Transcribed Image Text:Show that the function f(x, y) = xy – 6y2 is differentiable by finding values of ɛ, and ɛ, that satisfy the following definition.
Definition: If z = f(x, y), then fis differentiable at (a, b) if Az can be expressed in the form Az = f,(a, b) Ax + f,(a, b) Ay + 8,Ax + 8,Ay, where &, and s, →
(Дх, Ду) — (0, 0).
To find an expression for Az, we choose an arbitrary point (a, b), another point arbitrarily close to the first (a + Ax, b + Ay), and compute the difference between their respective
f-values.
Дz 3 f(a + Дх, b + Ду) - f(a, ь) %3D (а + Дх) (b + Ду) - 6(b + ду)? -
— ab + aдy + ьдх + ДхДу - бь? - 12bДу - 6(Ду)2- ab +
- (L
+ bAx + AxAy -
but f,(a, b) =
and f,(a, b) =
, so
Az = f,(a, b)Ax + f,(a, b)Ay + AXAY – 6AyAy.
Let ɛ, = Ay and ɛ, = -6Ay, therefore f is differentiable by the given definition.
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