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

1. compute all of the first order partial derivatives for the following func- tions. simplify when reasonable.\n(a) (f(x,y,z)=x^{2}y^{3}z^{4}+x^{5}y - pi xz^{6}+sqrt{7}y^{8}z).\n(b) (g(x,y)=sin(xe^{x^{2}y})).\n(c) (h(x,y)=\frac{x - y}{sqrt{x^{2}+y^{2}}}).\n(d) (f(x,y,z)=x^{2}cos(xy + 2xz-3yz)).\n(e) (g(x,y)=e^{(\frac{x^{2}-y^{2}}{x^{4}+y^{4}})}).\n(f) (h(x,y)=y^{2}ln(1 + xe^{x^{2}y^{3}})).\n(g) (phi(x,y)=\tan(x - y)).\n(h) (phi(w,x,y,z)=sin(wy^{3}-x^{2}z^{4})e^{w - x}).

1. compute all of the first order partial derivatives for the following func- tions. simplify when reasonable.\n(a) (f(x,y,z)=x^{2}y^{3}z^{4}+x^{5}y - pi xz^{6}+sqrt{7}y^{8}z).\n(b) (g(x,y)=sin(xe^{x^{2}y})).\n(c) (h(x,y)=\frac{x - y}{sqrt{x^{2}+y^{2}}}).\n(d) (f(x,y,z)=x^{2}cos(xy + 2xz-3yz)).\n(e) (g(x,y)=e^{(\frac{x^{2}-y^{2}}{x^{4}+y^{4}})}).\n(f) (h(x,y)=y^{2}ln(1 + xe^{x^{2}y^{3}})).\n(g) (phi(x,y)=\tan(x - y)).\n(h) (phi(w,x,y,z)=sin(wy^{3}-x^{2}z^{4})e^{w - x}).

Answer

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 rules like the power rule $\frac{\partial}{\partial x}(x^n)=nx^{n - 1}$, product rule $\frac{\partial(uv)}{\partial x}=u\frac{\partial v}{\partial x}+v\frac{\partial u}{\partial x}$, chain rule $\frac{\partial f(g(x))}{\partial x}=f^\prime(g(x))g^\prime(x)$.

(a)

  1. Find $\frac{\partial f}{\partial x}$:
    • $\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(x^{2}y^{3}z^{4}+x^{5}y-\pi xz^{6}+\sqrt{7}y^{8}z)$
    • Using the power rule, $\frac{\partial}{\partial x}(x^{2}y^{3}z^{4}) = 2xy^{3}z^{4}$, $\frac{\partial}{\partial x}(x^{5}y)=5x^{4}y$, $\frac{\partial}{\partial x}(-\pi xz^{6})=-\pi z^{6}$, and $\frac{\partial}{\partial x}(\sqrt{7}y^{8}z) = 0$.
    • So, $\frac{\partial f}{\partial x}=2xy^{3}z^{4}+5x^{4}y-\pi z^{6}$.
  2. Find $\frac{\partial f}{\partial y}$:
    • $\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(x^{2}y^{3}z^{4}+x^{5}y-\pi xz^{6}+\sqrt{7}y^{8}z)$
    • $\frac{\partial}{\partial y}(x^{2}y^{3}z^{4})=3x^{2}y^{2}z^{4}$, $\frac{\partial}{\partial y}(x^{5}y)=x^{5}$, $\frac{\partial}{\partial y}(-\pi xz^{6}) = 0$, $\frac{\partial}{\partial y}(\sqrt{7}y^{8}z)=8\sqrt{7}y^{7}z$.
    • So, $\frac{\partial f}{\partial y}=3x^{2}y^{2}z^{4}+x^{5}+8\sqrt{7}y^{7}z$.
  3. Find $\frac{\partial f}{\partial z}$:
    • $\frac{\partial f}{\partial z}=\frac{\partial}{\partial z}(x^{2}y^{3}z^{4}+x^{5}y-\pi xz^{6}+\sqrt{7}y^{8}z)$
    • $\frac{\partial}{\partial z}(x^{2}y^{3}z^{4}) = 4x^{2}y^{3}z^{3}$, $\frac{\partial}{\partial z}(x^{5}y)=0$, $\frac{\partial}{\partial z}(-\pi xz^{6})=-6\pi xz^{5}$, $\frac{\partial}{\partial z}(\sqrt{7}y^{8}z)=\sqrt{7}y^{8}$.
    • So, $\frac{\partial f}{\partial z}=4x^{2}y^{3}z^{3}-6\pi xz^{5}+\sqrt{7}y^{8}$.

(b)

  1. Find $\frac{\partial g}{\partial x}$:
    • Let $u = xe^{x^{2}y}$. Then $g(x,y)=\sin(u)$.
    • First, find $\frac{\partial u}{\partial x}$ using the product rule: $\frac{\partial u}{\partial x}=e^{x^{2}y}+x\cdot e^{x^{2}y}\cdot2xy=(1 + 2x^{2}y)e^{x^{2}y}$.
    • By the chain - rule, $\frac{\partial g}{\partial x}=\cos(xe^{x^{2}y})\cdot(1 + 2x^{2}y)e^{x^{2}y}$.
  2. Find $\frac{\partial g}{\partial y}$:
    • $\frac{\partial u}{\partial y}=x\cdot e^{x^{2}y}\cdot x^{2}=x^{3}e^{x^{2}y}$.
    • By the chain - rule, $\frac{\partial g}{\partial y}=\cos(xe^{x^{2}y})\cdot x^{3}e^{x^{2}y}$.

(c)

  1. Find $\frac{\partial h}{\partial x}$:
    • Using the quotient rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=x - y$, $v=\sqrt{x^{2}+y^{2}}=(x^{2}+y^{2})^{\frac{1}{2}}$.
    • $u^\prime = 1$, $v^\prime=\frac{2x}{2\sqrt{x^{2}+y^{2}}}=\frac{x}{\sqrt{x^{2}+y^{2}}}$.
    • $\frac{\partial h}{\partial x}=\frac{\sqrt{x^{2}+y^{2}}-(x - y)\frac{x}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{x^{2}+y^{2}-x^{2}+xy}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{y^{2}+xy}{(x^{2}+y^{2})^{\frac{3}{2}}}$.
  2. Find $\frac{\partial h}{\partial y}$:
    • $u^\prime=-1$, $v^\prime=\frac{2y}{2\sqrt{x^{2}+y^{2}}}=\frac{y}{\sqrt{x^{2}+y^{2}}}$.
    • $\frac{\partial h}{\partial y}=\frac{-\sqrt{x^{2}+y^{2}}-(x - y)\frac{y}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{-x^{2}-y^{2}-xy + y^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}=\frac{-x^{2}-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}$.

(d)

  1. Find $\frac{\partial F}{\partial x}$:
    • Using the product rule and chain rule. Let $u = x^{2}$, $v=\cos(xy + 2xz-3yz)$.
    • $\frac{\partial u}{\partial x}=2x$, $\frac{\partial v}{\partial x}=-\sin(xy + 2xz - 3yz)\cdot(y + 2z)$.
    • $\frac{\partial F}{\partial x}=2x\cos(xy + 2xz-3yz)-x^{2}(y + 2z)\sin(xy + 2xz - 3yz)$.
  2. Find $\frac{\partial F}{\partial y}$:
    • $\frac{\partial F}{\partial y}=-x^{2}\sin(xy + 2xz - 3yz)\cdot(x-3z)$.
  3. Find $\frac{\partial F}{\partial z}$:
    • $\frac{\partial F}{\partial z}=-x^{2}\sin(xy + 2xz - 3yz)\cdot(2x - 3y)$.

(e)

  1. Find $\frac{\partial G}{\partial x}$:
    • Let $u=\frac{x^{2}-y^{2}}{x^{4}+y^{4}}$. Then $G(x,y)=e^{u}$.
    • Using the quotient rule, $\frac{\partial u}{\partial x}=\frac{2x(x^{4}+y^{4})-(x^{2}-y^{2})\cdot4x^{3}}{(x^{4}+y^{4})^{2}}=\frac{2x^{5}+2xy^{4}-4x^{5}+4x^{3}y^{2}}{(x^{4}+y^{4})^{2}}=\frac{2xy^{4}-2x^{5}+4x^{3}y^{2}}{(x^{4}+y^{4})^{2}}$.
    • By the chain - rule, $\frac{\partial G}{\partial x}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\cdot\frac{2xy^{4}-2x^{5}+4x^{3}y^{2}}{(x^{4}+y^{4})^{2}}$.
  2. Find $\frac{\partial G}{\partial y}$:
    • $\frac{\partial u}{\partial y}=\frac{-2y(x^{4}+y^{4})-(x^{2}-y^{2})\cdot4y^{3}}{(x^{4}+y^{4})^{2}}=\frac{-2x^{4}y-2y^{5}-4x^{2}y^{3}+4y^{5}}{(x^{4}+y^{4})^{2}}=\frac{-2x^{4}y - 4x^{2}y^{3}+2y^{5}}{(x^{4}+y^{4})^{2}}$.
    • By the chain - rule, $\frac{\partial G}{\partial y}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\cdot\frac{-2x^{4}y - 4x^{2}y^{3}+2y^{5}}{(x^{4}+y^{4})^{2}}$.

(f)

  1. Find $\frac{\partial H}{\partial x}$:
    • Let $u = 1+xe^{x^{2}y^{3}}$. Then $H(x,y)=y^{2}\ln(u)$.
    • $\frac{\partial u}{\partial x}=e^{x^{2}y^{3}}+x\cdot e^{x^{2}y^{3}}\cdot2xy^{3}=(1 + 2x^{2}y^{3})e^{x^{2}y^{3}}$.
    • $\frac{\partial H}{\partial x}=\frac{y^{2}(1 + 2x^{2}y^{3})e^{x^{2}y^{3}}}{1+xe^{x^{2}y^{3}}}$.
  2. Find $\frac{\partial H}{\partial y}$:
    • $\frac{\partial u}{\partial y}=x\cdot e^{x^{2}y^{3}}\cdot3x^{2}y^{2}=3x^{3}y^{2}e^{x^{2}y^{3}}$.
    • $\frac{\partial H}{\partial y}=2y\ln(1 + xe^{x^{2}y^{3}})+\frac{y^{2}\cdot3x^{3}y^{2}e^{x^{2}y^{3}}}{1+xe^{x^{2}y^{3}}}$.

(g)

  1. Find $\frac{\partial\phi}{\partial x}$:
    • $\frac{\partial\phi}{\partial x}=\sec^{2}(x - y)$.
  2. Find $\frac{\partial\phi}{\partial y}$:
    • $\frac{\partial\phi}{\partial y}=-\sec^{2}(x - y)$.

(h)

  1. Find $\frac{\partial\Phi}{\partial w}$:
    • Using the product rule and chain rule. Let $u=\sin(wy^{3}-x^{2}z^{4})$ and $v = e^{w - x}$.
    • $\frac{\partial u}{\partial w}=y^{3}\cos(wy^{3}-x^{2}z^{4})$, $\frac{\partial v}{\partial w}=e^{w - x}$.
    • $\frac{\partial\Phi}{\partial w}=y^{3}\cos(wy^{3}-x^{2}z^{4})e^{w - x}+\sin(wy^{3}-x^{2}z^{4})e^{w - x}=e^{w - x}(y^{3}\cos(wy^{3}-x^{2}z^{4})+\sin(wy^{3}-x^{2}z^{4}))$.
  2. Find $\frac{\partial\Phi}{\partial x}$:
    • $\frac{\partial u}{\partial x}=-2xz^{4}\cos(wy^{3}-x^{2}z^{4})$, $\frac{\partial v}{\partial x}=-e^{w - x}$.
    • $\frac{\partial\Phi}{\partial x}=-2xz^{4}\cos(wy^{3}-x^{2}z^{4})e^{w - x}-\sin(wy^{3}-x^{2}z^{4})e^{w - x}=-e^{w - x}(2xz^{4}\cos(wy^{3}-x^{2}z^{4})+\sin(wy^{3}-x^{2}z^{4}))$.
  3. Find $\frac{\partial\Phi}{\partial y}$:
    • $\frac{\partial u}{\partial y}=3wy^{2}\cos(wy^{3}-x^{2}z^{4})$.
    • $\frac{\partial\Phi}{\partial y}=3wy^{2}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$.
  4. Find $\frac{\partial\Phi}{\partial z}$:
    • $\frac{\partial u}{\partial z}=-4x^{2}z^{3}\cos(wy^{3}-x^{2}z^{4})$.
    • $\frac{\partial\Phi}{\partial z}=-4x^{2}z^{3}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$.

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)

$\frac{\partial g}{\partial x}=\cos(xe^{x^{2}y})\cdot(1 + 2x^{2}y)e^{x^{2}y}$, $\frac{\partial g}{\partial y}=\cos(xe^{x^{2}y})\cdot x^{3}e^{x^{2}y}$

(c)

$\frac{\partial h}{\partial x}=\frac{y^{2}+xy}{(x^{2}+y^{2})^{\frac{3}{2}}}$, $\frac{\partial h}{\partial y}=\frac{-x^{2}-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}$

(d)

$\frac{\partial F}{\partial x}=2x\cos(xy + 2xz-3yz)-x^{2}(y + 2z)\sin(xy + 2xz - 3yz)$, $\frac{\partial F}{\partial y}=-x^{2}\sin(xy + 2xz - 3yz)\cdot(x-3z)$, $\frac{\partial F}{\partial z}=-x^{2}\sin(xy + 2xz - 3yz)\cdot(2x - 3y)$

(e)

$\frac{\partial G}{\partial x}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\cdot\frac{2xy^{4}-2x^{5}+4x^{3}y^{2}}{(x^{4}+y^{4})^{2}}$, $\frac{\partial G}{\partial y}=e^{\frac{x^{2}-y^{2}}{x^{4}+y^{4}}}\cdot\frac{-2x^{4}y - 4x^{2}y^{3}+2y^{5}}{(x^{4}+y^{4})^{2}}$

(f)

$\frac{\partial H}{\partial x}=\frac{y^{2}(1 + 2x^{2}y^{3})e^{x^{2}y^{3}}}{1+xe^{x^{2}y^{3}}}$, $\frac{\partial H}{\partial y}=2y\ln(1 + xe^{x^{2}y^{3}})+\frac{y^{2}\cdot3x^{3}y^{2}e^{x^{2}y^{3}}}{1+xe^{x^{2}y^{3}}}$

(g)

$\frac{\partial\phi}{\partial x}=\sec^{2}(x - y)$, $\frac{\partial\phi}{\partial y}=-\sec^{2}(x - y)$

(h)

$\frac{\partial\Phi}{\partial w}=e^{w - x}(y^{3}\cos(wy^{3}-x^{2}z^{4})+\sin(wy^{3}-x^{2}z^{4}))$, $\frac{\partial\Phi}{\partial x}=-e^{w - x}(2xz^{4}\cos(wy^{3}-x^{2}z^{4})+\sin(wy^{3}-x^{2}z^{4}))$, $\frac{\partial\Phi}{\partial y}=3wy^{2}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$, $\frac{\partial\Phi}{\partial z}=-4x^{2}z^{3}\cos(wy^{3}-x^{2}z^{4})e^{w - x}$