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Chapter 13: Problem 46

Find the first partial derivatives of the following functions. $$G(r, s, t)=\sqrt{r s+r t+s t}$$

Short Answer

Expert verified
Answer: The first partial derivatives of the function G(r, s, t) are: $$\frac{\partial G}{\partial r} = \frac{s + t}{2\sqrt{r s+r t+s t}}$$ $$\frac{\partial G}{\partial s} = \frac{r + t}{2\sqrt{r s+r t+s t}}$$ $$\frac{\partial G}{\partial t} = \frac{r + s}{2\sqrt{r s+r t+s t}}$$

Step by step solution

01

Rewrite the function in a more convenient form

Let's rewrite G(r, s, t) as: $$G(r, s, t) = (r s+r t+s t)^{1/2}$$
02

Determine the partial derivative with respect to r

Apply the chain rule, and differentiate G(r, s, t) with respect to r: $$\frac{\partial G}{\partial r} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot \frac{\partial}{\partial r}(r s+r t+s t)$$ Now differentiate the inside of the parentheses with respect to r: $$\frac{\partial}{\partial r}(r s+r t+s t) = s + t$$ Plug this back into the derivative: $$\frac{\partial G}{\partial r} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot (s + t)$$
03

Determine the partial derivative with respect to s

Following the same procedure, differentiate G(r, s, t) with respect to s: $$\frac{\partial G}{\partial s} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot \frac{\partial}{\partial s}(r s+r t+s t)$$ Now differentiate the inside of the parentheses with respect to s: $$\frac{\partial}{\partial s}(r s+r t+s t) = r + t$$ Plug this back into the derivative: $$\frac{\partial G}{\partial s} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot (r + t)$$
04

Determine the partial derivative with respect to t

Lastly, differentiate G(r, s, t) with respect to t: $$\frac{\partial G}{\partial t} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot \frac{\partial}{\partial t}(r s+r t+s t)$$ Now differentiate the inside of the parentheses with respect to t: $$\frac{\partial}{\partial t}(r s+r t+s t) = r+ s$$ Plug this back into the derivative: $$\frac{\partial G}{\partial t} = \frac{1}{2}(r s+r t+s t)^{-1/2} \cdot (r + s)$$
05

Give final results for the partial derivatives

The first partial derivatives of the function G(r, s, t) are: $$\frac{\partial G}{\partial r} = \frac{s + t}{2\sqrt{r s+r t+s t}}$$ $$\frac{\partial G}{\partial s} = \frac{r + t}{2\sqrt{r s+r t+s t}}$$ $$\frac{\partial G}{\partial t} = \frac{r + s}{2\sqrt{r s+r t+s t}}$$

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Most popular questions from this chapter

Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Identify and briefly describe the surfaces defined by the following equations. $$9 x^{2}+y^{2}-4 z^{2}+2 y=0$$

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y)=\ln \left(1+x^{2}+y^{2}\right)$$

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