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Chapter 3: Problem 48

Implicit differentiation with rational exponents Determine the slope of the following curves at the given point. $$(x+y)^{2 / 3}=y ;(4,4)$$

Short Answer

Expert verified
Answer: The slope of the curve at the point (4, 4) is -1.

Step by step solution

01

Implicit differentiation

Using the chain rule, implicitly differentiate the equation with respect to x: $$\frac{d}{dx}((x + y)^{2/3}) = \frac{d}{dx}(y)$$
02

Apply the chain rule

Apply the chain rule to \((x+y)^{2/3}\). This gives: $$\frac{2}{3}(x+y)^{-1/3}\frac{d}{dx}(x+y) = \frac{dy}{dx}$$
03

Add the derivative of y

Notice that \(\frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}\). Substituting this into the previous expression, we get: $$\frac{2}{3}(x+y)^{-1/3}(1+\frac{dy}{dx}) = \frac{dy}{dx}$$
04

Solve for the derivative of y

Now we need to solve for \(\frac{dy}{dx}\): $$\frac{2}{3}(x+y)^{-1/3}(1+\frac{dy}{dx}) - \frac{dy}{dx} = 0$$ Factor out \(\frac{dy}{dx}\): $$\frac{dy}{dx} \left[1 - \frac{2}{3}(x+y)^{-1/3} \right]=- \frac{2}{3}(x+y)^{-1/3}$$ Divide both sides by \(1 - \frac{2}{3}(x+y)^{-1/3}\): $$\frac{dy}{dx} = \frac{-\frac{2}{3}(x+y)^{-1/3}}{1 - \frac{2}{3}(x+y)^{-1/3}}$$
05

Evaluate the derivative at the given point

Substitute the values of the given point (4, 4) into the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{-\frac{2}{3}(4+4)^{-1/3}}{1 - \frac{2}{3}(4+4)^{-1/3}}$$ Simplify the expression: $$\frac{dy}{dx} = \frac{-\frac{2}{3}(8)^{-1/3}}{1 - \frac{2}{3}(8)^{-1/3}}$$ Calculate the final value: $$\frac{dy}{dx} = \frac{-\frac{2}{3}(2)^{-1}}{1 - \frac{2}{3}(2)^{-1}} = \frac{-\frac{2}{3}}{1 - \frac{2}{3}} = -1$$ Thus, the slope of the curve \((x+y)^{2 / 3}=y\) at the point (4, 4) is -1.

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Most popular questions from this chapter

Derivatives and inverse functions Suppose the slope of the curve \(y=f^{-1}(x)\) at (4,7) is \(\frac{4}{5}\) Find \(f^{\prime}(7)\)

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}(x f(x))\right|_{x=3}$$

Derivatives from tangent lines Suppose the line tangent to the graph of \(f\) at \(x=2\) is \(y=4 x+1\) and suppose \(y=3 x-2\) is the line tangent to the graph of \(g\) at \(x=2 .\) Find an equation of the line tangent to the following curves at \(x=2\) a. \(y=f(x) g(x)\) b. \(y=\frac{f(x)}{g(x)}\)

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of \(f^{-1}\) . $$f(x)=-x^{2}+8 ;(7,1)$$

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{f(x)}{(x+2)}\right]\right|_{x=4}$$
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