Questions

  1. Find partial derivative $\partial f/\partial x$ and $\partial f/\partial y$ of the following function

    • $f(x,y) = \ln(xy)\tan(xy)$.
    • $f(x,y) = u^2-v^2$, $u=\sec(x)$, and $v=\tan(y)$.

  2. Let $F(x,y)$ be a differtiable function. Let $F_x(x,y)=G(x,y)$ and $F_y(x,y)=H(x,y)$.

    • Find $u$ and $v$ such that $F(u,v) = F(x^3+1,2x)=f(x)$.
    • Use chain rule to find $f’(x)$, and express the answer in terms of $G$ and $H$.
    • Use the same reasoning to find the derivative of $f(x)=F(x,x)$, and express the derivative in terms of $G$ and $H$.

  3. The gradient $\nabla F(a,b)$ of a function $F(x,y)$ at $(a,b)$ is a vector defined by $(F_x(a,b), F_y(a,b))$. The gradient of a function can utilize to find the directional derivative $F_{(u,v)}$ of $F$ along a vector $(u,v)$. It is [F_{(u,v)}(a,b)=\nabla F(a,b)\cdot \frac{(u,v)}{|(u,v)|}] where $|(u,v)|=\sqrt{u^2+v^2}$ is the length of the vector $(u,v)$. Use the above introduction to calculate

    • The gradient of $F(x,y)=x^2+y^2$ at $(1,1)$ and $(0,0)$.
    • Observe that how the length of the gradient changes with respect to the point.
    • Find the directional derivative of $F$ at $(1,1)$ along the vector (3,4).
    • Confirm that the largest directional derivative of $F$ at $(1,1)$ is along the vector $(1,1)$.

Answers

    • $f_x=\frac{1}{x}\tan(xy) + \ln(xy)\sec^2(xy)y$, $f_y=\frac{1}{y}\tan(xy) + \ln(xy)\sec^2(xy)x$.
    • $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\ = 2(\sec(x))\sec(x)\tan(x) - 2(\tan(x))\cdot 0 = 2\sec^2(x)\tan(x).$ Similarly, $\frac{\partial f}{\partial y} = -2\tan(y)\sec^2(y)$.
    • $u=x^3+1$ and $v=2x$.
    • $f’(x) = \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} = G(x^3+1,2x)\cdot 3x^2 + H(x^3+1,2x)\cdot 2$.
    • We set $u=x$ and $v=x$, so $f’(x) = G(x,x) + H(x,x)$.
  3. The partial derivatives of $F$ are $\partial F/\partial x = 2x$ and $\partial F/\partial y=2y$.
    • $\nabla F(1,1) = (2,2)$.
    • $\nabla F(1,1)\frac{\vec{(3,4)}}{|(3,4)|} = \left(\frac{6}{5},\frac{8}{5}\right)$.
    • The largest directional derivative of $F$ at $(1,1)$ is attained along any vector that points to the same direction as the gradient $(2,2)$, including $(1,1)$.
    • The length of the gradient at any point $(a,b)$ is $|\nabla F(a,b)| = 2\sqrt{a^2+b^2} = 2|(a,b)-(0,0)|$. We observe that the length of the gradient is twice the distance between the point and the origin.