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Chapter 3: Problem 46

Implicit differentiation with rational exponents Determine the slope of the following curves at the given point. $$x^{2 / 3}+y^{2 / 3}=2 ;(1,1)$$

Short Answer

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

Step by step solution

01

Find the implicit derivative

Using the implicit differentiation method, we will find \(\frac{dy}{dx}\). First, we will differentiate both sides of the given equation with respect to \(x\). Differentiating \(x^{2/3}+y^{2/3}=2\) with respect to \(x\), we get: \(\frac{d}{dx}\left(x^{2/3}\right)+\frac{d}{dx}\left(y^{2/3}\right) = \frac{d}{dx}(2)\)
02

Apply the differentiation rules

We apply the power rule for differentiation and the chain rule for \(y^{2 / 3}\): \(\frac{2}{3}x^{-1/3}+\frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0\)
03

Solve for \(\frac{dy}{dx}\)

Now, we will isolate \(\frac{dy}{dx}\): \(\frac{2}{3}y^{-1/3}\frac{dy}{dx} = -\frac{2}{3}x^{-1/3}\) \(\frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \cdot \frac{3}{2}y^{1/3}\) \(\frac{dy}{dx} = -x^{-1/3}y^{1/3}\)
04

Substitute the given point values

We have found the implicit derivative of the equation. Now, we will substitute the coordinates of the given point \((1,1)\) into the \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -1^{-1/3}1^{1/3}\) \(\frac{dy}{dx} = -1\) So, the slope of the curve \(x^{2/3} + y^{2/3} = 2\) at the point \((1,1)\) is \(-1\).

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Most popular questions from this chapter

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

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{x}{x+2}\)

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Prove that \(\frac{d^{2 n}}{d x^{2 n}}(\sin x)=(-1)^{n} \sin x\) and \(\frac{d^{2 n}}{d x^{2 n}}(\cos x)=(-1)^{n} \cos x\)
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