A major corporation has branch offices in eight

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genyjoannerubiera

02.03.2022

The equation of the tangent line is $y = x + 0.5$ ====== $x^2 + y^2 = (2x^2+2y^2-x)^2$ the derivative at a point is the slope of the tangent line to that point.to implicitly differentiate, we assume that $y$ is a function of $x$ (so we need to apply chain rule when differentiating y terms)differentiate both sides $\dfrac{\text{d}}{\text{d}x}\left(x^2 + y^2\right) = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2$ for the left-hand side, use sum rule for differentiation: differentiate each term $\begin{aligned} \dfrac{\text{d}}{\text{d}x}\left(x^2\right) + \dfrac{\text{d}}{\text{d}x}\left(y^2\right) & = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \\ 2x + 2y \tfrac{\text{d}y}{\text{d}x}& = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \end{aligned}$ for the right-hand side, we need to use the chain rule.i will deal with right-hand side separately here, and then add it back to original equation after $\begin{aligned} \text{rhs} & = \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right)^2 \\ & = 2\left(2x^2+2y^2-x\right)^1 \cdot \dfrac{\text{d}}{\text{d}x}\left(2x^2+2y^2-x\right) \\ & = 2\left(2x^2+2y^2-x\right)\left(4x+4y\tfrac{\text{d}y}{\text{d}x}-1\right) \end{aligned}$ let's just connect this together now $\begin{aligned} 2x + 2y \tfrac{\text{d}y}{\text{d}x} & = 2\left(2x^2+2y^2-x\right)\left(4x+4y\tfrac{\text{d}y}{\text{d}x}-1\right) \end{aligned}$ we do not want to simplify this. you can if you want, but it involves a lot of time and unnecessary algebra. you might make mistakes. i might go over the character limit for these answers (around 5000 characters). stuff like that. we do not need to solve for $\text{d}y/\text{d}x$ in terms of $x$ and $y$ when we want to find the derivative at a point for an implicit curve.since we are interested in the point $(0,0.5)$ , substitute in 0 for all $x$ and 0.5 for all $y$ . then solve for $\text{d}y/\text{d}x$ , which is the slope of the tangent line.that x-value of 0 will really us out. $\begin{aligned} 2(0) + 2(0.5) \tfrac{\text{d}y}{\text{d}x} & = 2\left(2(0)^2+2(0.5)^2-0\right)\left(4(0)+4(0.5)\tfrac{\text{d}y}{\text{d}x}-1\right) \\ 1 \tfrac{\text{d}y}{\text{d}x} & = 2\big(2(0.25)\big)\left(2\tfrac{\text{d}y}{\text{d}x}-1\right) \\ \tfrac{\text{d}y}{\text{d}x} & = 2(0.5)\left(2\tfrac{\text{d}y}{\text{d}x}-1\right) \\ \tfrac{\text{d}y}{\text{d}x} & = 2\tfrac{\text{d}y}{\text{d}x}-1 \\ 0 & = \tfrac{\text{d}y}{\text{d}x} - 1 \\ 1 & = \tfrac{\text{d}y}{\text{d}x} \end{aligned}$ since $\text{d}y/\text{d}x$ is 1 at the point $(0,0.5)$ , the slope of the tangent line at point $(0,0.5)$ is 1.therefore, the slope of the line using point-slope form is $\begin{aligned} y - y_1 & = m(x-x_1) \\ y - 0.5 & = 1(x-0) \\ y & = x + 0.5 \end{aligned}$

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A major corporation has branch offices in eight major cities in the United States and Canada. The company president wants to visit five of these offices. How many different trip combinations are possible?

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