5.21 Derivative of composition functions

Theorem

Let \(f(x,y)=e^{x}\sin y\) where \(x(t)=t^2,y(t)=\sqrt{t}\), compute \(f'(t)\)
1. One way: Let's take the composition: \(g(t)=e^{t^2}\sin(\sqrt{t})\), then \(g^{\prime}(t)=e^{t^2}\cdot 2t\cdot \sin(\sqrt{t})+e^{t^2}\cos(\sqrt{t})\cdot \frac{1}{2} \frac{1}{\sqrt{t}}\)
2. Another way: let's \(\alpha(t)=(x(t),y(t))=(t^{2},\sqrt{t})\), then \(f \circ \alpha: \mathbb{R}\longrightarrow \mathbb{R}\)
  Then \(g(t)=f(\alpha(t))=f(x(t),y(t))\)
  \(g^{\prime}(t_{0})=\frac{\partial f}{\partial x}(x(t_{0}),y(t_{0}))x^{\prime}(t_{0} )+\frac{\partial f}{\partial y}(x(t_{0}),y(t_{0}))y^{\prime}(t_{0})\)
  
  \(\frac{\partial f}{\partial x}=e^{x}\sin y,\quad\frac{\partial f}{\partial y}=e^{x} \cos y\) and \(x'(t) = 2t, \quad y'(t) = \frac{1}{2\sqrt{t}}\)
  \(g'(t) = e^{t^2}\sin(\sqrt{t}) \cdot 2t + e^{t^2}\cos(\sqrt{t}) \cdot \frac{1}{2\sqrt{t}}\)
Compute the second order partial derivatives: \(f(x,y)=x^{3}y+e^{xy}+\sin(x+y)\)

\(f_{x}=3yx^{2}+ye^{xy}+\cos(x+y)\), \(f_{y}=x^{3}+xe^{xy}+\cos(x+y)\)

\(f_{xx}=6yx+y^2e^{xy}-\sin(x+y)\), \(f_{yy}=x^2e^{xy}-\sin(x+y)\), \(f_{yx}=3x^2+e^{xy}+xye^{xy}-\sin(x+y)\), \(f_{xy}=3x^2+e^{xy}+xye^{xy}-\sin(x+y)\)
\(f(x,y)= \begin{cases} x^{2}y^{2}\sin\left(\frac{1}{x^2+y^2}\right) & \text{if }(x,y)\neq (0,0) \\ 0 & \text{if }(x,y)=(0,0) \end{cases}\)

If \((x,y)\neq(0,0)\), then \(f_{x}=2y^{2}x\sin\left(\frac{1}{x^{2}+y^{2}}\right)+y^{2}x^{2}\cos\left(\frac{1}{x^{2}+y^{2}} \right)\cdot\left(-\frac{2x}{\left(x^{2}+y^{2}\right)^{2}}\right)\)

\(f_{y}=2x^{2}y\sin\left(\frac{1}{x^{2}+y^{2}}\right)+x^{2}y^{2}\cos\left(\frac{1}{x^{2}+y^{2}} \right)\cdot\left(-\frac{2y}{\left(x^{2}+y^{2}\right)^{2}}\right)\)

If \((x,y)=(0,0)\), then \(f_{x}\left(0,0\right)=\lim_{x\to0}\frac{f\left(x,0\right)-f\left(0,0\right)}{x}= \lim_{x\to0}\frac{x^{2}0^{2}\sin\left(\frac{1}{x^2+0^2}\right)-0}{x}=0\), \(f_y(0,0)=0\)

Thus \(f_{x}(x,y)= \begin{cases} 2y^{2}x\sin\left(\frac{1}{x^2+y^2}\right)+y^{2}x^{2}\cos\left(\frac{1}{x^2+y^2}\right)\cdot\left(-\frac{2x}{\left(x^2+y^2\right)^2}\right) & \text{if }(x,y)\neq(0,0) \\ 0 & \text{if }(x,y)=(0,0) \end{cases}\)

Easily, \(f_{xy}(0,0)=0\), in the same way, \(f_{yx}(0,0)=0\)

By symmetry, \(f_{xy}=f_{yx}\) for all points
\(f(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}\). Compute \(f_{xx}\) and \(f_{yx}\) at \((0,0)\).
At \((x,y) \neq (0,0)\), \(f_x(x,y) = \frac{(y(x^2-y^2)+2x^2y)(x^2+y^2) - 2x(xy(x^2-y^2))}{(x^2+y^2)^2} = \frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\)
Similar, \(f_y(x,y) = \frac{x(x^4-4x^2y^2-y^4)}{(x^2+y^2)^2}\)

At \((0,0)\), \(f_x(0,0) = \lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = \lim_{x \to 0} \frac{0-0}{x} = 0\) and \(f_y(0,0) = 0\)

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

Now. \(f_{xy}(0,0) = \lim_{y \to 0} \frac{f_x(0,y) - f_x(0,0)}{y-0}\) \(= \lim_{y \to 0} \frac{y(-y^4) - 0}{y^4}\) \(= \lim_{y \to 0} \frac{-y^5}{y^5} = -1\)

\(f_{yx}(0,0) = \lim_{x \to 0} \frac{f_y(x,0) - f_y(0,0)}{x-0}\) \(= \lim_{x \to 0} \frac{x(x^4) - 0}{x^4}\) \(= \lim_{x \to 0} \frac{x^5}{x^5} = 1\)

\(f_{xy}(0,0) = -1\), \(f_{yx}(0,0) = 1\)

Since the mixed partial derivatives are not continuous
Compute \(\displaystyle\iint_{R}\frac{x}{1+xy}dA\) where \(R=[0,1]\times [0,2]\)

\({\displaystyle\iint_{R}\frac{x}{1+xy}dA=\int_0^1\left(\int_0^2\frac{x}{1+xy}\,dy\right)dx}\)

Let \(1+xy=u,xdy=du\), then \(\displaystyle=\int_{0}^{1}\left(\int_{1}^{1+2x}\frac{1}{u}\,du\right)dx=\int^1_0\ln(1+2x)dx\)

Let \(1+2x=t,2dx=dt\), then \(\displaystyle=\frac{1}{2}\int^{3}_{1}\ln(t)dt\)

Let \(\ln(t)=u,u'=\frac{1}{t}\) and \(v'=1,v=t\), then \(\displaystyle =\frac{1}{2}\left(\ln(t)\cdot t\Big|^{3}_{1}-\int^{3}_{1}1\right)=\frac{3}{2}\ln (3)-1\)
Find the volume of the solid bounded below by the XY-plane and above by the surface \(z=e^{x+y}\) over the rectangle \(R = [0, 1] \times [0, \ln(2)]\).

\(\displaystyle\iint_{R}e^{x+y}dA=\int_{0}^{1}\left(\int_{0}^{\ln(2)}e^{x+y}\,dy\right )dx=\int^{1}_{0}\left(e^{x+y}\Big|^{\ln(2)}_{0}\right)dx=\int^{1}_{0}\left(e^{\ln(2)+x} -e^{x}\right)dx=e^{\ln(2)+1}-e-e^{\ln(2)}+1=2e-e-2+1=e-1\)
Find the volume of the region bounded below by the paraboloid \(z=x^2+y^2\) and above by the plane \(z=2x+4y+1\) over the rectangle \(R = [0,2]\times[0,1]\).

\(\displaystyle\iint_{R} x^2+y^2 dA=\int^{1}_{0}\left(\int^{2}_{0}(x^{2}+y^{2})dx\right )dy=\int^{1}_{0}\left(\frac{x^{3}}{3}+y^{2}x\right)\Big|^{2}_{0}dy=\int^{1}_{0}\left (\frac{8}{3}+2y^{2}\right)dy=\left(\frac{8}{3}y+\frac{2y^{3}}{3}\right)\Big|^{1}_{0} =\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

\(\displaystyle\iint_{R}2x+4y+1\,dA=\int^{1}_{0}\left(\int^{2}_{0}(2x+4y+1)dx\right )dy=\int^{1}_{0}\left(x^2+(4y+1)x\right)\Big|^{2}_{0}dy=\int^{1}_{0}\left (8y+6\right)dy=\left(4y^2+6y\right)\Big|^{1}_{0} =10\)

Then the result is \(\frac{20}{3}\)
Given the function \(f(x,y)\) and the region \(R\): \(f(x,y) = \frac{x^2 - y^2}{(x^2+y^2)^2}\) where \(R = [0,1] \times [0,1]\)

Compute the iterated integrals of \(f\) over \(R\).

\(\displaystyle\int_{0}^{1}\left( \int_{0}^{1}\frac{x^{2}- y^{2}}{(x^{2}+y^{2})^{2}} dy \right ) dx\), using partial fraction to solve the inner part

Then \(\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}=\frac{ay+b}{(x^{2}+y^{2})}+\frac{cy+d}{(x^{2}+y^{2})^{2}} =\frac{\left(ay+b\right)\left(x^{2}+y^{2}\right)+cy+d}{(x^{2}+y^{2})^{2}}=\frac{ayx^{2}+ay^{3}+bx^{2}+by^{2}+cy+d}{(x^{2}+y^{2})^{2}}\)

Then \(a=0,b=-1,c=0,d=2x^2\), then \(\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}=-\frac{1}{(x^{2}+y^{2})}+\frac{2x^{2}}{(x^{2}+y^{2})^{2}}\)

\(\displaystyle\int_0^1\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}dy=\int_0^1-\frac{1}{(x^{2}+y^{2})}+\frac{2x^{2}}{(x^{2}+y^{2})^{2}}dy=\int^1_0-\frac{1}{x^2+y^2}dy+\int^1_0\frac{2x^2}{(x^2+y^2)^2}dy=-\frac{1}{x}\arctan\left(\frac{y}{x}\right)+\frac{1}{x}\arctan\left(\frac{y}{x}\right)+\frac{y}{x^2+y^2}\Big|^1_0=\frac{1}{x^2+1}\)

Then \(\displaystyle\int_{0}^{1}\left( \int_{0}^{1}\frac{x^{2}- y^{2}}{(x^{2}+y^{2})^{2}} dy \right ) dx=\int^{1}_{0}\frac{1}{x^{2}+1}dx=\arctan(x)\Big|^{1}_{0}=\frac{\pi}{4}\)

And \({\displaystyle\int_0^1\left(\int_0^1\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}dx\right)dy=-\frac{\pi}{4}}\), we cannot use Fubini since \(f\) is not continuous at \((0,0)\)

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