HW-2: Derivatives & Chain Rule

The function \(f\) takes 3 inputs \((x, y, z)\) and returns a scalar, so its derivative \(f'\) is a \(1 \times 3\) Jacobian matrix containing the partial derivatives with respect to each variable: \[ f'(x,y,z) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{bmatrix} \]

We will use the chain rule.

1. Partial derivative with respect to \(x\) (\(\frac{\partial f}{\partial x}\)):
Treating \(y\) and \(z\) as constants: \begin{align*} \frac{\partial f}{\partial x} &= \cos(x \sin(y \sin z)) \cdot \frac{\partial}{\partial x}(x \sin(y \sin z)) \\ &= \cos(x \sin(y \sin z)) \cdot \sin(y \sin z) \end{align*}

2. Partial derivative with respect to \(y\) (\(\frac{\partial f}{\partial y}\)):
Treating \(x\) and \(z\) as constants: \begin{align*} \frac{\partial f}{\partial y} &= \cos(x \sin(y \sin z)) \cdot \frac{\partial}{\partial y}(x \sin(y \sin z)) \\ &= \cos(x \sin(y \sin z)) \cdot x \cos(y \sin z) \cdot \frac{\partial}{\partial y}(y \sin z) \\ &= \cos(x \sin(y \sin z)) \cdot x \cos(y \sin z) \cdot \sin z \end{align*}

3. Partial derivative with respect to \(z\) (\(\frac{\partial f}{\partial z}\)):
Treating \(x\) and \(y\) as constants: \begin{align*} \frac{\partial f}{\partial z} &= \cos(x \sin(y \sin z)) \cdot \frac{\partial}{\partial z}(x \sin(y \sin z)) \\ &= \cos(x \sin(y \sin z)) \cdot x \cos(y \sin z) \cdot \frac{\partial}{\partial z}(y \sin z) \\ &= \cos(x \sin(y \sin z)) \cdot x \cos(y \sin z) \cdot y \cos z \end{align*}

Combining these, the final answer is: \[ f'(x,y,z) = \begin{bmatrix} \sin(y \sin z) \cos(x \sin(y \sin z)) \\ x \sin z \cos(y \sin z) \cos(x \sin(y \sin z)) \\ x y \cos z \cos(y \sin z) \cos(x \sin(y \sin z)) \end{bmatrix}^T \]

We use the standard power rule \(d(u^n) = n u^{n-1} du\) along with the chain rule. The derivative will be a \(1 \times 2\) matrix.

1. Partial derivative with respect to \(x\): \begin{align*} \frac{\partial f}{\partial x} &= (\cos 3)[\sin(xy)]^{\cos 3 - 1} \cdot \frac{\partial}{\partial x}(\sin(xy)) \\ &= (\cos 3)[\sin(xy)]^{\cos 3 - 1} \cdot \cos(xy) \cdot y \\ &= y \cos 3 \cos(xy) [\sin(xy)]^{\cos 3 - 1} \end{align*}

2. Partial derivative with respect to \(y\): \begin{align*} \frac{\partial f}{\partial y} &= (\cos 3)[\sin(xy)]^{\cos 3 - 1} \cdot \frac{\partial}{\partial y}(\sin(xy)) \\ &= (\cos 3)[\sin(xy)]^{\cos 3 - 1} \cdot \cos(xy) \cdot x \\ &= x \cos 3 \cos(xy) [\sin(xy)]^{\cos 3 - 1} \end{align*}

The final answer is: \[ f'(x,y) = \begin{bmatrix} y \cos 3 \cos(xy) [\sin(xy)]^{\cos 3 - 1} & x \cos 3 \cos(xy) [\sin(xy)]^{\cos 3 - 1} \end{bmatrix} \]

This is a vector-valued function with 2 inputs and 3 outputs. Its derivative \(f'\) is a \(3 \times 2\) Jacobian matrix. Let component functions be: \(f_1(x,y) = \sin(xy)\)
\(f_2(x,y) = \sin(x \sin y)\)
\(f_3(x,y) = x^y\)

Finding partial derivatives for each component:

For \(f_1\):
\(\frac{\partial f_1}{\partial x} = y \cos(xy)\) and \(\frac{\partial f_1}{\partial y} = x \cos(xy)\)

For \(f_2\):
\(\frac{\partial f_2}{\partial x} = \sin y \cos(x \sin y)\) and \(\frac{\partial f_2}{\partial y} = x \cos y \cos(x \sin y)\)

For \(f_3\):
\(\frac{\partial f_3}{\partial x} = y x^{y-1}\) (power rule) and \(\frac{\partial f_3}{\partial y} = x^y \ln x\) (exponential rule)

Arranging these into the Jacobian matrix: \[ f'(x,y) = \begin{bmatrix} y \cos(xy) & x \cos(xy) \\ \sin y \cos(x \sin y) & x \cos y \cos(x \sin y) \\ y x^{y-1} & x^y \ln x \end{bmatrix} \]

We use the Fundamental Theorem of Calculus along with the Chain Rule. Let's rewrite the integral with a dummy variable \(t\): \(f(x,y) = \int_{a}^{xy} g(t) dt\).

1. Partial derivative with respect to \(x\) (treating \(y\) as constant): \[ \frac{\partial f}{\partial x} = g(xy) \cdot \frac{\partial}{\partial x}(xy) = g(xy) \cdot y \]

2. Partial derivative with respect to \(y\) (treating \(x\) as constant): \[ \frac{\partial f}{\partial y} = g(xy) \cdot \frac{\partial}{\partial y}(xy) = g(xy) \cdot x \]

The final derivative is a row matrix: \[ f'(x,y) = \begin{bmatrix} y \cdot g(xy) & x \cdot g(xy) \end{bmatrix} \]

A bilinear map on finite-dimensional spaces is always bounded. Thus, there exists a constant \(M > 0\) such that \(\vert f(h,k) \vert \le M \vert h \vert \vert k \vert\).

The norm of the vector \((h,k)\) is \(\vert (h,k) \vert = \sqrt{\vert h \vert^2 + \vert k \vert^2}\). This clearly means \(\vert h \vert \le \vert (h,k) \vert\).

We can write our limit expression as: \[ \frac{\vert f(h,k) \vert}{\vert (h,k) \vert} \le \frac{M \vert h \vert \vert k \vert}{\vert (h,k) \vert} = M \left( \frac{\vert h \vert}{\vert (h,k) \vert} \right) \vert k \vert \le M \cdot 1 \cdot \vert k \vert \]

As \((h,k) \to 0\), we have \(k \to 0\). So, \(M \vert k \vert \to 0\).
By the Squeeze Theorem, \(\lim_{(h,k) \to 0} \frac{\vert f(h,k) \vert}{\vert (h,k) \vert} = 0\). ∎

Let our candidate for the derivative be the linear map \(L(h,k) = f(a,k) + f(h,b)\). By the definition of the derivative, we need to show that: \[ \lim_{(h,k) \to 0} \frac{\vert f(a+h, b+k) - f(a,b) - L(h,k) \vert}{\vert (h,k) \vert} = 0 \]

Using bilinearity, we expand the first term: \[ f(a+h, b+k) = f(a,b) + f(a,k) + f(h,b) + f(h,k) \]

Substituting this into the numerator of our limit: \begin{align*} \text{Numerator} &= [f(a,b) + f(a,k) + f(h,b) + f(h,k)] - f(a,b) - [f(a,k) + f(h,b)] \\ &= f(h,k) \end{align*}

So the limit becomes \(\lim_{(h,k) \to 0} \frac{\vert f(h,k) \vert}{\vert (h,k) \vert}\). From Part (a), we know this limit is exactly \(0\).
Therefore, our candidate map is the correct derivative: \(Df(a,b)(x,y) = f(a,y) + f(x,b)\). ∎

The simple multiplication function \(p(x,y) = x \cdot y\) is bilinear because \(p(cx, y) = c(xy) = cp(x,y)\) and \(p(x_1+x_2, y) = (x_1+x_2)y = p(x_1, y) + p(x_2, y)\).

Since it is bilinear, we can use the formula from Part (b): \[ Dp(a,b)(x,y) = p(a,y) + p(x,b) = a \cdot y + x \cdot b = ay + bx \]

This is exactly the standard product rule for derivatives presented in Theorem 2-3. Thus, it is a special case. ∎

By the definition of an inverse function, for all \(x\), we have: \[ f(f^{-1}(x)) = x \]

We differentiate both sides with respect to \(x\).
For the left side, we use the multivariable Chain Rule. The derivative of the outer function \(f\) is evaluated at \(f^{-1}(x)\), and then multiplied by the derivative of the inner function \(f^{-1}(x)\): \[ \frac{d}{dx}[f(f^{-1}(x))] = f'(f^{-1}(x)) \cdot (f^{-1})'(x) \]

For the right side, the derivative of the identity function \(x\) is the Identity Matrix \(I\): \[ \frac{d}{dx}[x] = I \]

Equating both sides, we get: \[ f'(f^{-1}(x)) \cdot (f^{-1})'(x) = I \]

Now, substitute the specific point \(x = a\): \[ f'(f^{-1}(a)) \cdot (f^{-1})'(a) = I \]

Since the product of these two matrices is the Identity Matrix, they are inverses of each other. Multiplying both sides by the matrix inverse of \(f'(f^{-1}(a))\) gives: \[ (f^{-1})'(a) = [f'(f^{-1}(a))]^{-1} \] ∎

We use the chain rule. Let \(g'\) denote the derivative of the outer function \(g\).

1. Partial derivative with respect to \(x\): \[ \frac{\partial f}{\partial x} = g'(x+y) \cdot \frac{\partial}{\partial x}(x+y) = g'(x+y) \cdot 1 = g'(x+y) \]

2. Partial derivative with respect to \(y\): \[ \frac{\partial f}{\partial y} = g'(x+y) \cdot \frac{\partial}{\partial y}(x+y) = g'(x+y) \cdot 1 = g'(x+y) \]

Both partial derivatives are equal to \(g'(x+y)\).

Firstly, fix the variable \(x\) to some constant value \(x_0\). Define a single-variable function \(g(y) = f(x_0, y)\).
The derivative of \(g\) is the partial derivative of \(f\) with respect to \(y\): \[ g'(y) = \frac{\partial f}{\partial y}(x_0, y) = D_2f(x_0, y) \]

We are given that \(D_2f = 0\) everywhere. This means \(g'(y) = 0\) for all \(y\). By the Mean Value Theorem from single-variable calculus, if a function's derivative is zero everywhere, the function must be a constant.

Thus, \(f(x_0, y)\) does not change as \(y\) changes. Since \(x_0\) was arbitrary, this holds for any \(x\). Therefore, \(f\) is independent of \(y\). ∎

According to Part 1, since \(D_2f = 0\), the function \(f\) does not depend on \(y\). We can write it as a function of \(x\) alone: \(f(x,y) = h(x)\).

Now we use the second condition: \(D_1f = 0\).
The partial derivative of \(f\) with respect to \(x\) is just the regular derivative of \(h(x)\): \[ D_1f(x,y) = h'(x) = 0 \]

Since \(h'(x) = 0\) for all \(x\), the Mean Value Theorem again tells us that \(h(x)\) must be a constant \(C\).

Therefore, \(f(x,y) = h(x) = C\) for all pairs \((x,y)\). The function \(f\) is constant. ∎

We use the limit definition of partial derivatives. Note that \(f(x,0) = 0\) and \(f(0,y) = 0\).

1. Finding \(D_2f(x,0)\) (partial with respect to \(y\)): \[ D_2f(x,0) = \lim_{h \to 0} \frac{f(x,h) - f(x,0)}{h} = \lim_{h \to 0} \frac{xh \frac{x^2 - h^2}{x^2 + h^2} - 0}{h} \]

Cancel the \(h\): \[ D_2f(x,0) = \lim_{h \to 0} x \frac{x^2 - h^2}{x^2 + h^2} \]

As \(h \to 0\), the fraction becomes \(\frac{x^2}{x^2} = 1\). Thus, \(D_2f(x,0) = x(1) = x\).

2. Finding \(D_1f(0,y)\) (partial with respect to \(x\)): \[ D_1f(0,y) = \lim_{h \to 0} \frac{f(h,y) - f(0,y)}{h} = \lim_{h \to 0} \frac{hy \frac{h^2 - y^2}{h^2 + y^2} - 0}{h} \]

Cancel the \(h\): \[ D_1f(0,y) = \lim_{h \to 0} y \frac{h^2 - y^2}{h^2 + y^2} \]

As \(h \to 0\), the fraction becomes \(\frac{-y^2}{y^2} = -1\). Thus, \(D_1f(0,y) = y(-1) = -y\). ∎

1. For \(D_{1,2}f(0,0)\) (derivative of \(D_2f\) with respect to \(x\)): \[ D_{1,2}f(0,0) = \lim_{h \to 0} \frac{D_2f(h,0) - D_2f(0,0)}{h} \]

Using \(D_2f(x,0) = x\), we have \(D_2f(h,0) = h\) and \(D_2f(0,0) = 0\): \[ D_{1,2}f(0,0) = \lim_{h \to 0} \frac{h - 0}{h} = 1 \]

2. For \(D_{2,1}f(0,0)\) (derivative of \(D_1f\) with respect to \(y\)): \[ D_{2,1}f(0,0) = \lim_{h \to 0} \frac{D_1f(0,h) - D_1f(0,0)}{h} \]

Using \(D_1f(0,y) = -y\), we have \(D_1f(0,h) = -h\) and \(D_1f(0,0) = 0\): \[ D_{2,1}f(0,0) = \lim_{h \to 0} \frac{-h - 0}{h} = -1 \]

Clearly, \(1 \neq -1\), so \(D_{1,2}f(0,0) \neq D_{2,1}f(0,0)\). ∎

We will use mathematical induction and the limit definition of the derivative.

Step 1: Base Case (\(f'(0)\)) \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{e^{-h^{-2}}}{h} = \lim_{h \to 0} \frac{1/h}{e^{h^{-2}}} \]

Substitute \(t = 1/h\). As \(h \to 0\), \(t \to \pm \infty\). \[ f'(0) = \lim_{t \to \pm \infty} \frac{t}{e^{t^2}} \]

Using L'Hôpital's Rule, the derivative of the numerator is 1, and the denominator is \(2t e^{t^2}\): \[ f'(0) = \lim_{t \to \pm \infty} \frac{1}{2t e^{t^2}} = 0 \]

Step 2: General form for \(x \neq 0\)
Using the chain and product rules, every derivative \(f^{(n)}(x)\) for \(x \neq 0\) can be written as a product of a polynomial in terms of \((1/x)\) and the exponential function \(e^{-x^{-2}}\): \[ f^{(n)}(x) = P_n\left(\frac{1}{x}\right) e^{-x^{-2}} \] where \(P_n\) is some polynomial.

Step 3: Inductive Step
Assume that for the \(k\)-th derivative, \(f^{(k)}(0) = 0\). We must show \(f^{(k+1)}(0) = 0\). \[ f^{(k+1)}(0) = \lim_{h \to 0} \frac{f^{(k)}(h) - f^{(k)}(0)}{h} = \lim_{h \to 0} \frac{P_k(1/h) e^{-h^{-2}}}{h} \]

Again, substitute \(t = 1/h\): \[ f^{(k+1)}(0) = \lim_{t \to \pm \infty} \frac{t \cdot P_k(t)}{e^{t^2}} \]

The numerator is a new polynomial. Because the exponential function \(e^{t^2}\) grows much faster than any polynomial, repeated application of L'Hôpital's Rule will eventually reduce the numerator to a constant, while the exponential remains in the denominator. Therefore, the limit is \(0\). \[ f^{(k+1)}(0) = 0 \]

Since all derivatives exist and are equal to \(0\) at the origin, the function is \(C^\infty\). ∎