Question

HELP MEeeeeeeeee g: R² → R a differentiable function at (0, 0), with g (x, y) = 0 only at the point (x, y) = (0, 0). Consider
`f(x,y)\left \{ {{(tan^2(g(x,y)) if (x,y)\\eq (0,0))/(g(x,y) ) } \atop {0 if (x,y)=(0,0)}} \right.`

(a) Calculate as partial derivatives of f in (0, 0), in terms of the partial derivatives of g.

(b) Show that f is differentiable at (0, 0).

Answer 1

(a) This follows from the definition for the partial derivative, with the help of some limit properties and a well-known limit.

• Recall that for
`f:\mathbb R^2\to\mathbb R`, we have the partial derivative with respect to
`x` defined as

`\displaystyle (\partial f)/(\partial x) = \lim_(h\to0)\frac{f(x+h,y) - f(x,y)}h`

The derivative at (0, 0) is then

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)\frac{f(0+h,0) - f(0,0)}h`

• By definition of
`f`,
`f(0,0)=0`, so

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)\frac{f(h,0)}h = \lim_(h\to0)(\tan^2(g(h,0)))/(h\cdot g(h,0))`

• Expanding the tangent in terms of sine and cosine gives

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)(\sin^2(g(h,0)))/(h\cdot g(h,0) \cdot \cos^2(g(h,0)))`

• Introduce a factor of
`g(h,0)` in the numerator, then distribute the limit over the resulting product as

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)(\sin^2(g(h,0)))/(g(h,0)^2) \cdot \lim_(h\to0)\frac1{\cos^2(g(h,0))} \cdot \lim_(h\to0)\frac{g(h,0)}h`

• The first limit is 1; recall that for
`a\\eq0`, we have

`\displaystyle\lim_(x\to0)(\sin(ax))/(ax)=1`

The second limit is also 1, which should be obvious.

• In the remaining limit, we end up with

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)\frac{g(h,0)}h = \lim_(h\to0)\frac{g(h,0)-g(0,0)}h`

and this is exactly the partial derivative of
`g` with respect to
`x`.

`\displaystyle (\partial f)/(\partial x)(0,0) = \lim_(h\to0)\frac{g(h,0)-g(0,0)}h = (\partial g)/(\partial x)(0,0)`

For the same reasons shown above,

`\displaystyle (\partial f)/(\partial y)(0,0) = (\partial g)/(\partial y)(0,0)`

(b) To show that
`f` is differentiable at (0, 0), we first need to show that
`f` is continuous.

• By definition of continuity, we need to show that

`\left|f(x,y)-f(0,0)\right|`

is very small, and that as we move the point
`(x,y)` closer to the origin,
`f(x,y)` converges to
`f(0,0)`.

We have

`\left|f(x,y)-f(0,0)\right| = \left|(\tan^2(g(x,y)))/(g(x,y))\right| \\\\ = \left|(\sin^2(g(x,y)))/(g(x,y)^2)\cdot(g(x,y))/(\cos^2(g(x,y)))\right| \\\\ = \left|(\sin(g(x,y)))/(g(x,y))\right|^2 \cdot (|g(x,y)|)/(\cos^2(x,y))`

The first expression in the product is bounded above by 1, since
`|\sin(x)|\le|x|` for all
`x`. Then as
`(x,y)` approaches the origin,

`\displaystyle\lim_((x,y)\to(0,0))(|g(x,y)|)/(\cos^2(x,y)) = 0`

So,
`f` is continuous at the origin.

• Now that we have continuity established, we need to show that the derivative exists at (0, 0), which amounts to showing that the rate at which
`f(x,y)` changes as we move the point
`(x,y)` closer to the origin, given by

`\left|(f(x,y)-f(0,0))/(√(x^2+y^2))\right|,`

approaches 0.

Just like before,

`\left|(\tan^2(g(x,y)))/(g(x,y)√(x^2+y^2))\right| = \left|(\sin^2(g(x,y)))/(g(x,y))\right|^2 \cdot \left|(g(x,y))/(\cos^2(g(x,y))√(x^2+y^2))\right| \\\\ \le (|g(x,y)|)/(\cos^2(g(x,y))√(x^2+y^2))`

and this converges to
`g(0,0)=0`, since differentiability of
`g` means

`\displaystyle \lim_((x,y)\to(0,0))(g(x,y)-g(0,0))/(√(x^2+y^2))=0`

So,
`f` is differentiable at (0, 0).