HW-3: Directional Derivatives & Chain Rule

Let the intermediate variables be \(u(x,y) = g(x)k(y)\) and \(v(x,y) = g(x) + h(y)\). Thus, \(F(x,y) = f(u, v)\).

By the Chain Rule, the partial derivative with respect to \(x\) is: \[ \frac{\partial F}{\partial x} = D_1f \cdot \frac{\partial u}{\partial x} + D_2f \cdot \frac{\partial v}{\partial x} \]

Calculating the inner derivatives: \[ \frac{\partial u}{\partial x} = g'(x)k(y), \quad \frac{\partial v}{\partial x} = g'(x) \]

Substitute back: \[ \frac{\partial F}{\partial x} = D_1f \cdot g'(x)k(y) + D_2f \cdot g'(x) \]

For \(y\): \[ \frac{\partial F}{\partial y} = D_1f \cdot \frac{\partial u}{\partial y} + D_2f \cdot \frac{\partial v}{\partial y} \]

Inner derivatives: \[ \frac{\partial u}{\partial y} = g(x)k'(y), \quad \frac{\partial v}{\partial y} = h'(y) \]

Substitute back: \[ \frac{\partial F}{\partial y} = D_1f \cdot g(x)k'(y) + D_2f \cdot h'(y) \]

Let \(u(x,y,z) = g(x + y)\) and \(v(x,y,z) = h(y + z)\).

Inner derivatives for \(u\): \[ \frac{\partial u}{\partial x} = g'(x+y), \quad \frac{\partial u}{\partial y} = g'(x+y), \quad \frac{\partial u}{\partial z} = 0 \]

Inner derivatives for \(v\): \[ \frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = h'(y+z), \quad \frac{\partial v}{\partial z} = h'(y+z) \]

Applying the Multivariable Chain Rule: \begin{align*} \frac{\partial F}{\partial x} &= D_1f \cdot g'(x+y) + D_2f \cdot 0 = D_1f \cdot g'(x+y) \\ \frac{\partial F}{\partial y} &= D_1f \cdot g'(x+y) + D_2f \cdot h'(y+z) \\ \frac{\partial F}{\partial z} &= D_1f \cdot 0 + D_2f \cdot h'(y+z) = D_2f \cdot h'(y+z) \end{align*}

Let \(u(x,y,z) = x^y\), \(v(x,y,z) = y^z\), and \(w(x,y,z) = z^x\). Thus, \(F(x,y,z) = f(u,v,w)\).

Calculating the inner derivatives for \(x\): \[ \frac{\partial u}{\partial x} = y x^{y-1}, \quad \frac{\partial v}{\partial x} = 0, \quad \frac{\partial w}{\partial x} = z^x \ln z \]

Substitute back using the Chain Rule: \[ \frac{\partial F}{\partial x} = D_1f \cdot y x^{y-1} + D_2f \cdot 0 + D_3f \cdot z^x \ln z = D_1f \cdot y x^{y-1} + D_3f \cdot z^x \ln z \]

For \(y\): \[ \frac{\partial u}{\partial y} = x^y \ln x, \quad \frac{\partial v}{\partial y} = z y^{z-1}, \quad \frac{\partial w}{\partial y} = 0 \]

Substitute back: \[ \frac{\partial F}{\partial y} = D_1f \cdot x^y \ln x + D_2f \cdot z y^{z-1} + D_3f \cdot 0 = D_1f \cdot x^y \ln x + D_2f \cdot z y^{z-1} \]

For \(z\): \[ \frac{\partial u}{\partial z} = 0, \quad \frac{\partial v}{\partial z} = y^z \ln y, \quad \frac{\partial w}{\partial z} = x z^{x-1} \]

Substitute back: \[ \frac{\partial F}{\partial z} = D_1f \cdot 0 + D_2f \cdot y^z \ln y + D_3f \cdot x z^{x-1} = D_2f \cdot y^z \ln y + D_3f \cdot x z^{x-1} \]

Let \(u=x\), \(v=g(x)\), and \(w=h(x,y)\).

For \(x\): \[ \frac{\partial F}{\partial x} = D_1f \cdot (1) + D_2f \cdot g'(x) + D_3f \cdot \frac{\partial h}{\partial x}(x,y) \]

For \(y\): \[ \frac{\partial F}{\partial y} = D_1f \cdot (0) + D_2f \cdot (0) + D_3f \cdot \frac{\partial h}{\partial y}(x,y) = D_3f \cdot \frac{\partial h}{\partial y}(x,y) \]

By definition for standard basis vector \(e_i\): \[ D_{e_i} f(a) = \lim_{t \to 0} \frac{f(a + t e_i) - f(a)}{t} \]

Since \(a + t e_i = (a_1, \dots, a_i + t, \dots, a_n)\), we have: \[ \lim_{t \to 0} \frac{f(a_1, \dots, a_i + t, \dots, a_n) - f(a_1, \dots, a_n)}{t} \]

This is the fundamental definition of the partial derivative \(D_i f(a)\). ∎

\[ D_{cx} f(a) = \lim_{h \to 0} \frac{f(a + h(cx)) - f(a)}{h} \]

If \(c = 0\), both sides are \(0\). If \(c \neq 0\), let \(u = hc\). As \(h \to 0\), \(u \to 0\). Replacing \(h\) with \(u/c\): \[ \lim_{u \to 0} \frac{f(a + ux) - f(a)}{u/c} = c \left[ \lim_{u \to 0} \frac{f(a + ux) - f(a)}{u} \right] = c D_x f(a) \] ∎

By definition of differentiability at \(a\): \[ \lim_{v \to 0} \frac{f(a + v) - f(a) - Df(a)(v)}{|v|} = 0 \]

Let \(v = tx\). By linearity of \(Df(a)\), \(Df(a)(tx) = t Df(a)(x)\). Multiplying by \(|x|\): \[ \lim_{t \to 0} \frac{f(a + tx) - f(a) - t Df(a)(x)}{|t|} = 0 \implies \lim_{t \to 0} \frac{f(a + tx) - f(a)}{t} = Df(a)(x) \]

Thus, \(D_x f(a) = Df(a)(x)\). Because \(Df(a)\) is a linear transformation: \[ D_{x+y} f(a) = Df(a)(x+y) = Df(a)(x) + Df(a)(y) = D_x f(a) + D_y f(a) \] ∎

Part 1: For \(x \neq 0\) and \(t > 0\): \(f(tx) = t|x| g\left(\frac{x}{|x|}\right) = t f(x)\).

For \(t < 0\): \(f(tx) = -t|x| g\left(-\frac{x}{|x|}\right) = t|x| g\left(\frac{x}{|x|}\right) = t f(x)\).

\[ D_x f(0) = \lim_{t \to 0} \frac{f(tx)}{t} = \lim_{t \to 0} \frac{tf(x)}{t} = f(x) \]

Thus, \(D_x f(0)\) exists everywhere and equals \(f(x)\).

---

Part 2: If \(g \neq 0\), choose \(v = (v_1, v_2)\) where \(g(v) \neq 0\). Note \(v_1, v_2 \neq 0\).

Let \(x = (v_1, 0)\) and \(y = (0, v_2)\).

\(f(x) = |x| g(\pm(1,0)) = 0\) and \(f(y) = |y| g(\pm(0,1)) = 0\). Thus \(f(x) + f(y) = 0\).

However, \(f(x+y) = f(v) = |v|g(v) \neq 0\).

Thus \(f(x+y) \neq f(x) + f(y)\), breaking linearity. ∎

First Part: \(f(0,0) = 0\). Along the path \(y = \frac{1}{2}x^2\), \(0 < \frac{1}{2}x^2 < x^2\), so \(f(x, \frac{1}{2}x^2) = 1\).

\[ \lim_{x \to 0} f(x, \frac{1}{2}x^2) = 1 \neq f(0,0) \]

Therefore, \(f\) is discontinuous at \((0,0)\).

---

Part 2: For direction \(v = (v_1, v_2)\): \[ D_v f(0,0) = \lim_{t \to 0} \frac{f(tv_1, tv_2)}{t} \]

For \(v_2 \neq 0\), the condition for \(f=1\) is \(0 < tv_2 < t^2v_1^2\). As \(t \to 0\), \(t^2\) dominates, making the inequality false for small \(t\). Thus \(f(tv_1, tv_2) = 0\) near \(t=0\). The limit evaluates to \(0\). Directional derivatives exist everywhere and equal to \(0\). ∎

At \(x=0\), using the limit definition: \[ f'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h)}{h} = \lim_{h \to 0} h \sin(1/h) = 0 \quad \text{(by Squeeze Theorem)} \]

For \(x \neq 0\), the derivative is: \[ f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \]

As \(x \to 0\), \(2x \sin(1/x) \to 0\), but \(\cos(1/x)\) oscillates between -1 and 1. The limit \(\lim_{x \to 0} f'(x)\) does not exist, so \(f'\) is not continuous at \(0\). ∎

\(D_1 f(0,0) = \lim_{h \to 0} \frac{h^2 \sin(1/|h|)}{h} = 0\). By symmetry, \(D_2 f(0,0) = 0\).

The differentiability limit: \[ \lim_{(h,k) \to (0,0)} \frac{|f(h,k)|}{\sqrt{h^2+k^2}} = \lim_{(h,k) \to (0,0)} \sqrt{h^2+k^2} \left|\sin\frac{1}{\sqrt{h^2+k^2}}\right| = 0 \]

Thus, \(f\) is differentiable.

For \((x,y) \neq (0,0)\): \[ D_1 f(x,y) = 2x \sin\frac{1}{\sqrt{x^2 + y^2}} - \frac{x}{\sqrt{x^2 + y^2}} \cos\frac{1}{\sqrt{x^2 + y^2}} \]

Along \(y=0\), \(D_1 f(x,0) = 2x \sin(1/|x|) - \frac{x}{|x|} \cos(1/|x|)\). As \(x \to 0\), the second term oscillates indefinitely. Hence, \(D_1 f\) is discontinuous at \((0,0)\). ∎

Theorem 2-8 establishes that a function \(f: \mathbf{R}^n \to \mathbf{R}^m\) is differentiable at \(a\) if all its partial derivatives exist in an open set containing \(a\) and are continuous at \(a\). We want to show that \(D_1f\) only needs to exist at \(a\), without requiring continuity. Since differentiability of a vector-valued function is equivalent to the differentiability of its components, we consider \(f: \mathbf{R}^n \to \mathbf{R}\) without loss of generality.

To prove differentiability at \(a\), we must show that: \[ \lim_{h \to 0} \frac{f(a+h) - f(a) - \sum_{i=1}^n D_if(a)h_i}{|h|} = 0 \]

Consider the telescoping sum for \(f(a+h) - f(a)\), ordering the sequence of points so that the first coordinate is shifted first:

Let \(p_0 = a = (a_1, a_2, \dots, a_n)\).
\(p_1 = (a_1+h_1, a_2, \dots, a_n)\).
\(p_2 = (a_1+h_1, a_2+h_2, \dots, a_n)\).
\dots
\(p_n = a+h\).

We can write \(f(a+h) - f(a) = \sum_{i=1}^n (f(p_i) - f(p_{i-1}))\).

For the first term (\(i=1\)), by the fundamental definition of the partial derivative at \(a\): \[ f(p_1) - f(p_0) = f(a_1+h_1, a_2, \dots, a_n) - f(a) = D_1f(a)h_1 + r(h_1) \]
where \(\lim_{h_1 \to 0} \frac{r(h_1)}{h_1} = 0\). Note that this step only requires the existence of \(D_1f\) at \(a\).

For \(i \ge 2\), since \(D_if\) exists and is continuous in a neighborhood of \(a\), we apply the Mean Value Theorem on the segment between \(p_{i-1}\) and \(p_i\): \[ f(p_i) - f(p_{i-1}) = D_if(q_i)h_i \]
where \(q_i\) lies on the line segment between \(p_{i-1}\) and \(p_i\). As \(h \to 0\), \(q_i \to a\). By the continuity of \(D_if\) (for \(i \ge 2\)), we have \(D_if(q_i) \to D_if(a)\).

Substituting these expressions back into the limit fraction: \[ f(a+h) - f(a) - \sum_{i=1}^n D_if(a)h_i = r(h_1) + \sum_{i=2}^n (D_if(q_i) - D_if(a))h_i \]

Dividing by \(|h|\) and taking the absolute value: \[ \frac{\left|f(a+h) - f(a) - \sum_{i=1}^n D_if(a)h_i\right|}{|h|} \le \frac{|r(h_1)|}{|h|} + \sum_{i=2}^n |D_if(q_i) - D_if(a)| \frac{|h_i|}{|h|} \]

Since \(|h_1| \le |h|\) and \(|h_i| \le |h|\), we can bound this by: \[ \le \frac{|r(h_1)|}{|h_1|} + \sum_{i=2}^n |D_if(q_i) - D_if(a)| \]

As \(h \to 0\), the first term \(\frac{|r(h_1)|}{|h_1|} \to 0\) (due to the existence of \(D_1f(a)\)) and the second term \(|D_if(q_i) - D_if(a)| \to 0\) (due to the continuity of the remaining partial derivatives).

Therefore, the limit evaluates to \(0\), proving \(f\) is differentiable at \(a\) even without the continuity of \(D_1f\). ∎

Let \(g(t) = f(tx)\). Using the Chain Rule on \((tx^1, \dots, tx^n)\): \[ g'(t) = \sum_{i=1}^n D_i f(tx) \cdot x^i \]

Using the homogeneous property \(g(t) = t^m f(x)\): \[ g'(t) = m t^{m-1} f(x) \]

Equating both expressions and setting \(t=1\): \[ \sum_{i=1}^n x^i D_i f(x) = m (1)^{m-1} f(x) = m f(x) \] ∎

Let \(h_x(t) = f(tx)\). By the Fundamental Theorem of Calculus: \[ \int_0^1 h_x'(t) dt = h_x(1) - h_x(0) = f(x) - f(0) = f(x) \]

By the Chain Rule: \[ h_x'(t) = \sum_{i=1}^n D_i f(tx) \cdot x^i \]

Substitute into the integral: \[ f(x) = \int_0^1 \left( \sum_{i=1}^n x^i D_i f(tx) \right) dt = \sum_{i=1}^n x^i \left( \int_0^1 D_i f(tx) dt \right) \]

Defining \(g_i(x) = \int_0^1 D_i f(tx) dt\) yields \(f(x) = \sum_{i=1}^n x^i g_i(x)\). ∎

Assume, for contradiction, that \(f\) is 1-1. Since \(f\) is injective, it cannot be a constant function. Thus, there exists a point \((a,b) \in \mathbf{R}^2\) such that \(\nabla f(a,b) \neq 0\). Without loss of generality, assume \(D_1 f(a,b) \neq 0\).

Define \(g: \mathbf{R}^2 \to \mathbf{R}^2\) by \(g(x,y) = (f(x,y), y)\).
The Jacobian matrix of \(g\) is: \[ Dg(x,y) = \begin{pmatrix} D_1 f(x,y) & D_2 f(x,y) \\ 0 & 1 \end{pmatrix} \]

At \((a,b)\), \(\det Dg(a,b) = D_1 f(a,b) \neq 0\).

By the Inverse Function Theorem, there exists an open neighborhood \(U\) of \((a,b)\) and an open neighborhood \(V\) of \(g(a,b)\) such that \(g: U \to V\) is a diffeomorphism.

Since \(V\) is open, it contains distinct points of the form \((c, y_1)\) and \((c, y_2)\) where \(y_1 \neq y_2\).

Let \(u_1 = g^{-1}(c, y_1)\) and \(u_2 = g^{-1}(c, y_2)\) in \(U\). Since \((c, y_1) \neq (c, y_2)\), we have \(u_1 \neq u_2\).

However, by the definition of \(g\), \(g(x,y) = (f(x,y), y)\) implies that \(f(u_1) = c\) and \(f(u_2) = c\).

Therefore, \(f(u_1) = f(u_2)\) for \(u_1 \neq u_2\), which contradicts the assumption that \(f\) is 1-1. ∎

Let \(k\) be the maximal rank of \(Df\) on \(\mathbf{R}^n\). Note that \(k \le m < n\).

Suppose this maximal rank \(k\) is attained at a point \(p \in \mathbf{R}^n\). By permuting the coordinates if necessary, assume without loss of generality that the upper-left \(k \times k\) submatrix of \(Df(p)\) is invertible.

Define \(G: \mathbf{R}^n \to \mathbf{R}^n\) by: \[ G(x) = (f_1(x), \dots, f_k(x), x_{k+1}, \dots, x_n) \]

The Jacobian matrix \(DG(p)\) is block-triangular, and \(\det DG(p) \neq 0\).

By the Inverse Function Theorem, \(G\) is a local diffeomorphism near \(p\).

Consider the local change of coordinates \(y = G(x)\). In these coordinates, \(f\) takes the form: \[ \tilde{f}(y) = f(G^{-1}(y)) = (y_1, \dots, y_k, \tilde{f}_{k+1}(y), \dots, \tilde{f}_m(y)) \]

Since coordinate transformations do not change the rank, the rank of \(D\tilde{f}\) is at most \(k\). Because the first \(k\) components already provide a rank of \(k\), we must have \(\frac{\partial \tilde{f}_i}{\partial y_j} = 0\) for all \(i > k\) and \(j > k\).

Since \(n > m \ge k\), the variable \(y_n\) exists. The condition \(\frac{\partial \tilde{f}}{\partial y_n} = 0\) implies \(\tilde{f}\) is completely independent of \(y_n\).

Thus, for a small \(\epsilon > 0\), we have: \[ \tilde{f}(y_1, \dots, y_{n-1}, 0) = \tilde{f}(y_1, \dots, y_{n-1}, \epsilon) \]

This shows \(f\) maps distinct points to the exact same value. Therefore, \(f\) is not 1-1. ∎

Let \(x_1, x_2 \in \mathbf{R}\) with \(x_1 \neq x_2\). Without loss of generality, assume \(x_1 < x_2\).

Since \(f\) is differentiable everywhere on \(\mathbf{R}\), it is continuous on \([x_1, x_2]\) and differentiable on \((x_1, x_2)\).

By the Mean Value Theorem, there exists \(c \in (x_1, x_2)\) such that: \[ f(x_2) - f(x_1) = f'(c)(x_2 - x_1) \]

We are given that \(f'(a) \neq 0\) for all \(a \in \mathbf{R}\), which implies \(f'(c) \neq 0\).

Since \(x_1 < x_2\), we know \(x_2 - x_1 > 0\). Thus, the product \(f'(c)(x_2 - x_1) \neq 0\), which yields: \[ f(x_2) - f(x_1) \neq 0 \implies f(x_1) \neq f(x_2) \]

Since \(f(x_1) \neq f(x_2)\) for all \(x_1 \neq x_2\), the function \(f\) is 1-1. ∎

First, we compute the Jacobian matrix \(f'(x,y)\): \[ f'(x,y) = \begin{pmatrix} \frac{\partial}{\partial x}(e^x \cos y) & \frac{\partial}{\partial y}(e^x \cos y) \\ \frac{\partial}{\partial x}(e^x \sin y) & \frac{\partial}{\partial y}(e^x \sin y) \end{pmatrix} = \begin{pmatrix} e^x \cos y & -e^x \sin y \\ e^x \sin y & e^x \cos y \end{pmatrix} \]

The Jacobian determinant is: \[ \det f'(x,y) = (e^x \cos y)(e^x \cos y) - (-e^x \sin y)(e^x \sin y) \] \[ \det f'(x,y) = e^{2x} \cos^2 y + e^{2x} \sin^2 y = e^{2x}(\cos^2 y + \sin^2 y) \]

Using the identity \(\cos^2 y + \sin^2 y = 1\), we obtain: \[ \det f'(x,y) = e^{2x} \]

Since the exponential function is strictly positive for all \(x \in \mathbf{R}\), \(e^{2x} > 0\). Thus, \(\det f'(x,y) \neq 0\) for all \((x,y) \in \mathbf{R}^2\).

Next, we show \(f\) is not 1-1. Consider the distinct points \((0, 0)\) and \((0, 2\pi)\) in \(\mathbf{R}^2\): \[ f(0, 0) = (e^0 \cos 0, e^0 \sin 0) = (1, 0) \] \[ f(0, 2\pi) = (e^0 \cos 2\pi, e^0 \sin 2\pi) = (1, 0) \]

We observe that \(f(0, 0) = f(0, 2\pi)\) despite \((0, 0) \neq (0, 2\pi)\).

Since distinct inputs map to the same output, \(f\) is not injective. ∎