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5.14 Partial Derivatives

  1. \(f(x,y)=\cos(x^{4})\cdot e^{x^2y-5y^3}\), compute the first order partial derivatives of \(f\) at \((x,y)\)​

    \(\frac{\partial f}{\partial x}(x,y)=-\sin(x^{4})\cdot 4x^{3}\cdot e^{x^2y-5y^3}+\cos (x^{4})\cdot e^{x^2y-5y^3}\cdot2xy\)​

    \(\frac{\partial f}{\partial y}(x,y)=\cos(x^{4})\cdot e^{x^2y-5y^3}\cdot\left(x^{2} -15y^{2}\right)\)

  2. \(f(x,y)= \begin{cases} \frac{x^{3}+x^{4}-y^{3}}{x^{2}+y^{2}} & \text{if }(x,y)\neq (0,0) \\ 0 & \text{if }(x,y)=(0,0) \end{cases}\)

    If \((x,y)\neq (0,0)\), then \(\frac{\partial f}{\partial x}(x,y)=\frac{(3x^2+4x^3)(x^2+y^2)-(x^3+x^4-y^3)(2x)}{(x^2+y^2)^2}\)​

    \(\frac{\partial f}{\partial y}(x,y)=\frac{-3y^2(x^2+y^2)-(x^3+x^4-y^3)2y}{(x^2+y^2)^2}\)

    If \((x,y)=(0,0)\), then \(\frac{\partial f}{\partial x}(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\to0} \frac{f(x,0)}{x}=\lim_{x\to0}\frac{\frac{x^{3}+x^{4}}{x^{2}}}{x}=\lim_{x\to0}\frac{x^{3}+x^{4}}{x^{3}} =1\)​

    Then \(\frac{\partial f}{\partial y}(0,0)=-1\) in same way

  3. Check the differentiability of the following functions

    \(g(x,y)=\ln(1-x^{2}-y^{2})\) and \(\text{Domain}=\{(x,y)\in \R^2:x^2+y^2<1\}\)​

    By Theorem, \(\frac{\partial f}{\partial x}(x,y)=\frac{-2x}{1-x^2-y^2}\) and \(\frac{\partial f}{\partial y}(x,y)=\frac{-2y}{1-x^2-y^2}\)​

    Since both partial derivatives are continuous on \(D\), then \(g\) is differentiable on \(D\)​

    \(f(x,y)= \begin{cases} xy\sin\left(\frac{1}{x^2+y^2}\right) & \text{if }(x,y)\neq (0,0) \\ 0 & \text{if }(x,y)=(0,0) \end{cases}\)

    Let \((x,y) \neq (0,0)\)
    ​\(\frac{\partial f}{\partial x}(x,y) = y \sin\left(\frac{1}{x^2+y^2}\right) + xy \cos\left(\frac{1}{x^2+y^2}\right) \left(\frac{-2x}{(x^2+y^2)^2}\right)\)

    So \(\frac{\partial f}{\partial x}(x,y) = \begin{cases} y \sin\left(\frac{1}{x^2+y^2}\right) + xy \cos\left(\frac{1}{x^2+y^2}\right) \left(\frac{-2x}{(x^2+y^2)^2}\right) & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}\) is not continuous

    Thus we use the definition to prove it, at \((0,0)\)

    \(f_{x}(0,0)=\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x}=\lim_{x\to 0}\frac{x\cdot 0\cdot \sin(\frac{1}{x^2})-0}{x}=0\) and \(f_y(0,0)=0\)

    Let's see if \(\lim_{(x,y)\to(0,0)}\frac{f(x,y) - f(0,0) - 0}{||(x,y)||}= 0\)

    For \(\left| \frac{xy \sin\left(\frac{1}{x^2+y^2}\right)}{\sqrt{x^2+y^2}} \right|\) we need to show that this goes to 0 when \((x,y)\to(0,0)\)
    ​\(\le \left| \frac{xy}{\sqrt{x^2+y^2}} \right| \le \frac{|x||y|}{\sqrt{x^2}} = |y| \xrightarrow{(x,y)\to(0,0)} 0\)

    Then \(f\) is differentiable at \((0,0)\).

  4. Find the tangent plane to the surface \(z=x^2+3xy-y^2\) at the point \((1,2,3)\)

    Recall: \(z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\)​

    \(\frac{\partial f}{\partial x}(x,y)=2x+3y,\frac{\partial f}{\partial y}(x,y)=3x-2y\) are continuous, then \(f\) is differentiable

    Then \(z=f(1,2)+(2+3\cdot 2)(x-1)+(3-2\cdot 2)(y-2)=3+8x-8-y+2=8x-y-3\)

  5. Let \(v=(\cos \theta,\sin\theta)\) be unit vector for \(\theta\in [0,2\pi)\) and \(f(x,y)=x^2+y^2\)

    Recall Definition and Theorem

    \(\frac{\partial f}{\partial x}(x,y)=\lim_{t\to0}\frac{f\left(x+t\cos\theta,y+t\sin\theta\right)-f\left(x,y\right)}{t} =\lim_{t\to0}\frac{\left(x+t\cos\theta\right)^{2}+\left(y+t\sin\theta\right)^{2}-\left(x^{2}+y^{2}\right)}{t}\)​

    \(=\lim_{t\to0}t+2x\cos\theta+2y\sin\theta=2x\cos\theta+2y\sin\theta\). This exists for every direction

    In particular, if \(\theta=0,\frac{\partial f}{\partial v}=2x=\frac{\partial f}{\partial x}\text{ and }\theta=\frac{\pi}{2},\frac{\partial f}{\partial v}=2y=\frac{\partial f}{\partial y}\)​

    \(f(x,y)= \begin{cases} \frac{xy}{x^{2}+y^{2}} & \text{if }(x,y)\neq(0,0) \\ 0 & \text{if }(x,y)=(0,0) \end{cases}\) at \((0,0)\)

    \(\frac{\partial f}{\partial v}\left(0,0\right)=\lim_{t\to0}\frac{f\left(t\cos\theta,t\sin\theta\right)-f\left(0,0\right)}{t} =\lim_{t\to0}\frac{\frac{t^{2}\sin\theta\cos\theta}{t^{2}}}{t}=\lim_{t\to0}\frac{\sin\theta \cos\theta}{t}\)

    This limit exists if \(\theta = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2}\). Thus the function doesn't have any directional derivatives. It only has four directional derivatives

    \(f(x,y)=|x|+|y|\) at \((0,0)\)​

    \(\frac{\partial f}{\partial v}(0,0)=\lim_{t\to0}\frac{f(t\sin\theta,t\sin\theta)-f(0,0)}{t} =\lim_{t\to0}\frac{\left|t\sin\theta\right|+\left|t\sin\theta\right|}{t}=\lim_{t\to0} \frac{\left|t\right|}{t}\left(\left|\cos\theta|+|\sin\theta\right|\right)\)​

    This limit doesn't exist for every \(\theta\in [0,2\pi)\), \(f\) doesn't have directional derivatives at \((0,0)\)​

    Let \(f(x,y) = \begin{cases} \frac{x^2 y}{x^4+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}\)

    Compute the directional der. of \(f\) at \((0,0)\) in the direction of the unit vector \(v=(a,b)\).

    \(\frac{\partial f}{\partial v}(0,0) = \lim_{t \to 0} \frac{f(ta, tb) - f(0,0)}{t} = \lim_{t \to 0} \frac{\frac{(ta)^2 (tb)}{(ta)^4 + (tb)^2} - 0}{t}\) \(= \lim_{t \to 0} \frac{\frac{t^2 a^2 tb}{t^4 a^4 + t^2 b^2}}{t} = \lim_{t \to 0} \frac{\frac{t^3 a^2 b}{t^2 (t^2 a^4 + b^2)}}{t}\)
    \(= \lim_{t \to 0} \frac{t^3 a^2 b}{t^3 (t^2 a^4 + b^2)} = \lim_{t \to 0} \frac{a^2 b}{t^2 a^4 + b^2}\)

    If \(b=0\), \(\frac{\partial f}{\partial v}(0,0) = 0\), assume \(b \neq 0\), so \(\lim_{t \to 0} \frac{a^2 b}{t^2 a^4 + b^2} = \frac{a^2 b}{0 \cdot a^4 + b^2} = \frac{a^2 b}{b^2} = \frac{a^2}{b}\)

    Then \(f\) has directional derivatives at \((0,0)\) for every direction.

    (Note that \(f\) is not differentiable. Infact, it's not even continuous)