1. compute all of the first order partial derivatives for the following func - tions. simplify when…

Explanation:

Step1: Recall partial - derivative rules

When finding $\frac{\partial f}{\partial x}$, treat $y$ and $z$ as constants, and vice - versa for $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$. Use the power rule $\frac{\partial}{\partial x}(x^n)=nx^{n - 1}$, product rule $\frac{\partial(uv)}{\partial x}=u\frac{\partial v}{\partial x}+v\frac{\partial u}{\partial x}$, and chain rule $\frac{\partial f(g(x))}{\partial x}=f^\prime(g(x))g^\prime(x)$.

Step2: Find partial derivatives of $f(x,y,z)=x^{2}y^{3}z^{4}+x^{5}y-\pi xz^{6}+\sqrt{7}y^{8}z$

• $\frac{\partial f}{\partial x}=2xy^{3}z^{4}+5x^{4}y-\pi z^{6}$ (Treat $y$ and $z$ as constants and apply power rule)
• $\frac{\partial f}{\partial y}=3x^{2}y^{2}z^{4}+x^{5}+8\sqrt{7}y^{7}z$ (Treat $x$ and $z$ as constants and apply power rule)
• $\frac{\partial f}{\partial z}=4x^{2}y^{3}z^{3}-6\pi xz^{5}+\sqrt{7}y^{8}$ (Treat $x$ and $y$ as constants and apply power rule)

Step3: Find partial derivatives of $g(x,y)=\sin(xe^{x^{2}y})$

Let $u = xe^{x^{2}y}$. Then $\frac{\partial u}{\partial x}=e^{x^{2}y}+2x^{2}ye^{x^{2}y}$ (Product rule and chain rule for $e^{x^{2}y}$) and $\frac{\partial u}{\partial y}=x^{3}e^{x^{2}y}$ (Treat $x$ as constant and chain rule for $e^{x^{2}y}$).

• $\frac{\partial g}{\partial x}=\cos(xe^{x^{2}y})(e^{x^{2}y}+2x^{2}ye^{x^{2}y})$ (Chain rule $\frac{\partial g}{\partial x}=\cos(u)\frac{\partial u}{\partial x}$)
• $\frac{\partial g}{\partial y}=\cos(xe^{x^{2}y})(x^{3}e^{x^{2}y})$ (Chain rule $\frac{\partial g}{\partial y}=\cos(u)\frac{\partial u}{\partial y}$)

Step4: Find partial derivatives of $h(x,y)=\frac{x - y}{\sqrt{x^{2}+y^{2}}}$

Rewrite as $h(x,y)=(x - y)(x^{2}+y^{2})^{-\frac{1}{2}}$. Using product rule $\frac{\partial h}{\partial x}=(x^{2}+y^{2})^{-\frac{1}{2}}+(x - y)\left(-\frac{1}{2}\right)(x^{2}+y^{2})^{-\frac{3}{2}}(2x)=\frac{x^{2}+y^{2}-x(x - y)}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{y(x + y)}{(x^{2}+y^{2})^{\frac{3}{2}}}$ Using product rule $\frac{\partial h}{\partial y}=-(x^{2}+y^{2})^{-\frac{1}{2}}+(x - y)\left(-\frac{1}{2}\right)(x^{2}+y^{2})^{-\frac{3}{2}}(2y)=\frac{-(x^{2}+y^{2})-y(x - y)}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{-x(x + y)}{(x^{2}+y^{2})^{\frac{3}{2}}}$

Step5: Find partial derivatives of $F(x,y,z)=x^{2}\cos(xy + 2xz-3yz)$

Let $u=xy + 2xz-3yz$. Then $\frac{\partial u}{\partial x}=y + 2z$, $\frac{\partial u}{\partial y}=x-3z$, $\frac{\partial u}{\partial z}=2x-3y$.

• $\frac{\partial F}{\partial x}=2x\cos(xy + 2xz - 3yz)-x^{2}(y + 2z)\sin(xy + 2xz - 3yz)$ (Product rule and chain rule)
• $\frac{\partial F}{\partial y}=-x^{2}(x - 3z)\sin(xy + 2xz - 3yz)$ (Chain rule)
• $\frac{\partial F}{\partial z}=-x^{2}(2x - 3y)\sin(xy + 2xz - 3yz)$ (Chain rule)

Step6: Find partial derivatives of $G(x,y)=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}$

Let $u=\frac{x^{2}-y^{2}}{x^{4}+y^{4}}$. Then $\frac{\partial u}{\partial x}=\frac{2x(x^{4}+y^{4})-(x^{2}-y^{2})(4x^{3})}{(x^{4}+y^{4})^{2}}=\frac{2xy^{4}-2x^{5}}{(x^{4}+y^{4})^{2}}$ and $\frac{\partial u}{\partial y}=\frac{-2y(x^{4}+y^{4})-(x^{2}-y^{2})(4y^{3})}{(x^{4}+y^{4})^{2}}=\frac{-2x^{4}y - 2y^{5}}{(x^{4}+y^{4})^{2}}$.

• $\frac{\partial G}{\partial x}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\frac{2xy^{4}-2x^{5}}{(x^{4}+y^{4})^{2}}$ (Chain rule)
• $\frac{\partial G}{\partial y}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\frac{-2x^{4}y - 2y^{5}}{(x^{4}+y^{4})^{2}}$ (Chain rule)

Step7: Find partial derivatives of $H(x,y)=y^{2}\ln(1 + xe^{x^{2}y^{3}})$

Let $u = 1+xe^{x^{2}y^{3}}$. Then $\frac{\partial u}{\partial x}=e^{x^{2}y^{3}}+2x^{2}y^{3}e^{x^{2}y^{3}}$ and $\frac{\partial u}{\partial y}=3x^{3}y^{2}e^{x^{2}y^{3}}$.

• $\frac{\partial H}{\partial x}=\frac{y^{2}(e^{x^{2}y^{3}}+2x^{2}y^{3}e^{x^{2}y^{3}})}{1 + xe^{x^{2}y^{3}}}$ (Product rule and chain rule)
• $\frac{\partial H}{\partial y}=2y\ln(1 + xe^{x^{2}y^{3}})+\frac{3x^{3}y^{4}e^{x^{2}y^{3}}}{1 + xe^{x^{2}y^{3}}}$ (Product rule and chain rule)

Step8: Find partial derivatives of $\phi(x,y)=\tan(x - y)$

• $\frac{\partial\phi}{\partial x}=\sec^{2}(x - y)$ (Chain rule, derivative of $\tan(u)$ with $u=x - y$ and $\frac{\partial u}{\partial x}=1$)
• $\frac{\partial\phi}{\partial y}=-\sec^{2}(x - y)$ (Chain rule, derivative of $\tan(u)$ with $u=x - y$ and $\frac{\partial u}{\partial y}=-1$)

Step9: Find partial derivatives of $\Phi(w,x,y,z)=\sin(wy^{3}-x^{2}z^{4})e^{w - x}$

Let $u = wy^{3}-x^{2}z^{4}$ and $v=w - x$.

• $\frac{\partial\Phi}{\partial w}=\cos(wy^{3}-x^{2}z^{4})y^{3}e^{w - x}+\sin(wy^{3}-x^{2}z^{4})e^{w - x}=e^{w - x}(\cos(wy^{3}-x^{2}z^{4})y^{3}+\sin(wy^{3}-x^{2}z^{4}))$ (Product rule and chain rule)
• $\frac{\partial\Phi}{\partial x}=-2xz^{4}\cos(wy^{3}-x^{2}z^{4})e^{w - x}-\sin(wy^{3}-x^{2}z^{4})e^{w - x}=e^{w - x}(-2xz^{4}\cos(wy^{3}-x^{2}z^{4})-\sin(wy^{3}-x^{2}z^{4}))$ (Product rule and chain rule)
• $\frac{\partial\Phi}{\partial y}=3wy^{2}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$ (Chain rule)
• $\frac{\partial\Phi}{\partial z}=-4x^{2}z^{3}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$ (Chain rule)

Answer:

(a) $\frac{\partial f}{\partial x}=2xy^{3}z^{4}+5x^{4}y-\pi z^{6}$, $\frac{\partial f}{\partial y}=3x^{2}y^{2}z^{4}+x^{5}+8\sqrt{7}y^{7}z$, $\frac{\partial f}{\partial z}=4x^{2}y^{3}z^{3}-6\pi xz^{5}+\sqrt{7}y^{8}$ (b) $\frac{\partial g}{\partial x}=\cos(xe^{x^{2}y})(e^{x^{2}y}+2x^{2}ye^{x^{2}y})$, $\frac{\partial g}{\partial y}=\cos(xe^{x^{2}y})(x^{3}e^{x^{2}y})$ (c) $\frac{\partial h}{\partial x}=\frac{y(x + y)}{(x^{2}+y^{2})^{\frac{3}{2}}}$, $\frac{\partial h}{\partial y}=\frac{-x(x + y)}{(x^{2}+y^{2})^{\frac{3}{2}}}$ (d) $\frac{\partial F}{\partial x}=2x\cos(xy + 2xz - 3yz)-x^{2}(y + 2z)\sin(xy + 2xz - 3yz)$, $\frac{\partial F}{\partial y}=-x^{2}(x - 3z)\sin(xy + 2xz - 3yz)$, $\frac{\partial F}{\partial z}=-x^{2}(2x - 3y)\sin(xy + 2xz - 3yz)$ (e) $\frac{\partial G}{\partial x}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\frac{2xy^{4}-2x^{5}}{(x^{4}+y^{4})^{2}}$, $\frac{\partial G}{\partial y}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\frac{-2x^{4}y - 2y^{5}}{(x^{4}+y^{4})^{2}}$ (f) $\frac{\partial H}{\partial x}=\frac{y^{2}(e^{x^{2}y^{3}}+2x^{2}y^{3}e^{x^{2}y^{3}})}{1 + xe^{x^{2}y^{3}}}$, $\frac{\partial H}{\partial y}=2y\ln(1 + xe^{x^{2}y^{3}})+\frac{3x^{3}y^{4}e^{x^{2}y^{3}}}{1 + xe^{x^{2}y^{3}}}$ (g) $\frac{\partial\phi}{\partial x}=\sec^{2}(x - y)$, $\frac{\partial\phi}{\partial y}=-\sec^{2}(x - y)$ (h) $\frac{\partial\Phi}{\partial w}=e^{w - x}(\cos(wy^{3}-x^{2}z^{4})y^{3}+\sin(wy^{3}-x^{2}z^{4}))$, $\frac{\partial\Phi}{\partial x}=e^{w - x}(-2xz^{4}\cos(wy^{3}-x^{2}z^{4})-\sin(wy^{3}-x^{2}z^{4}))$, $\frac{\partial\Phi}{\partial y}=3wy^{2}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$, $\frac{\partial\Phi}{\partial z}=-4x^{2}z^{3}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$