# 微积分作业代写calclulus代考| Partial derivatives

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## 微积分作业代写calclulus代考|Mastery Check 2.5:

Let $f(x, y, z)=x y+z \sin (y z)$. Using Definition 2.3, determine $\frac{\partial f}{\partial y}$ at an arbitrary point $(x, y, z)$.
Hint: You may need to use standard limits (see Page 24).
$c$
In solving this Mastery Check problem you will have noticed that you could
2.C Partial derivatives
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have and would have arrived at the same result had you used the rules of differentiation for functions of one variable, provided you treated $x$ and $z$ as if they were constants! In actual fact, Definition $2.3$ effectively states that in taking the limit with respect to one variable, we do keep all other variables fixed. It should not come as a surprise that we find this equivalence. We demonstrate this very convenient operational equivalence with an example and leave it to Mastery Check $2.6$ to reinforce the procedure.

## 微积分作业代写calclulus代考|Mastery Check 2.6:

Find the (first-order) partial derivatives of the following functions with respect to each variable:

1. $f(x, y, z)=\frac{x^{2}+y^{2}}{x^{2}-z^{2}}$
2. $f(x, y, u, v)=x^{2} \sin (2 y) \ln (2 u+3 v)$;
3. $f(s, t, u)=\sqrt{s^{2} t+s t u+t u^{2}}$;
4. $f(x, y, z)=y \sin ^{-1}\left(x^{2}-z^{2}\right)$;
5. $f(x, y, z, u)=\sin (3 x) \cosh (2 y)-\cos (3 z) \sinh (2 u)$
6. $f(u, v, w)=u^{2} \mathrm{e}^{u v^{2} w}$
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Differentiation
Now that we can evaluate them, what are partial derivatives?
Let’s look more closely at Figure 2.3 (Page 52). Given the foregoing discussion and particularly Definition $2.3$ we consider two specific cases of that graph of the function of two variables.

C

2.C 偏导数
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## 微积分作业代写calclulus代考|Mastery Check 2.6:

1. F(X,和,和)=X2+和2X2−和2
2. F(X,和,你,v)=X2没有⁡(2和)ln⁡(2你+3v);
3. F(s,吨,你)=s2吨+s吨你+吨你2;
4. F(X,和,和)=和没有−1⁡(X2−和2);
5. F(X,和,和,你)=没有⁡(3X)科什⁡(2和)−某物⁡(3和)出生⁡(2你)
6. F(你,v,在)=你2和你v2在
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微分
现在我们可以评估它们，什么是偏导数？
让我们更仔细地看一下图 2.3（第 52 页）。鉴于上述讨论，特别是定义2.3我们考虑两个变量的函数图的两个特定情况。