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Use implicit differentiation to find \(\frac{d y}{d x}\) $$e^{x y}=2 y$$

Short Answer

Expert verified
Question: Find the derivative of the function \(y(x)\) with respect to \(x\) given the implicit equation \(e^{x y}=2 y\). Answer: The derivative of the function \(y(x)\) with respect to \(x\) is \(\frac{dy}{dx} = \frac{e^{xy}y}{e^{xy}x - 2}\).

Step by step solution

01

Differentiate both sides of the equation with respect to x

Start by differentiating the left side with respect to \(x\). Use the chain rule to differentiate the \(e^{xy}\) term: $$\frac{d}{dx}(e^{xy}) = e^{xy}\frac{d}{dx}(xy)$$ Now, differentiate the right side of the equation with respect to \(x\). Remember to include the \(\frac{dy}{dx}\) term when differentiating \(y\): $$\frac{d}{dx}(2y) = 2\frac{dy}{dx}$$ We now have the differentiated equation: $$e^{xy}\frac{d}{dx}(xy) = 2\frac{dy}{dx}$$
02

Differentiate the product xy with respect to x

We now need to differentiate the product \(xy\) with respect to \(x\). Use the product rule: $$\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$$ Now, substitute this result back into the differentiated equation from Step 1: $$e^{xy}(x\frac{dy}{dx} + y) = 2\frac{dy}{dx}$$
03

Isolate the dy/dx term

We now need to isolate the \(\frac{dy}{dx}\) term in the equation. Start by expanding the equation: $$e^{xy}x\frac{dy}{dx} + e^{xy}y = 2\frac{dy}{dx}$$ Now, move the term with \(\frac{dy}{dx}\) on the right side to the left side: $$e^{xy}x\frac{dy}{dx} - 2\frac{dy}{dx} = e^{xy}y$$ Factor out the \(\frac{dy}{dx}\) term from the left side: $$\frac{dy}{dx}(e^{xy}x - 2) = e^{xy}y$$ Finally, divide both sides by \((e^{xy}x - 2)\) to isolate the \(\frac{dy}{dx}\) term: $$\frac{dy}{dx} = \frac{e^{xy}y}{e^{xy}x - 2}$$
04

Final Answer

We have successfully isolated the \(\frac{dy}{dx}\) term and found the derivative of the function \(y(x)\) with respect to \(x\): $$\frac{dy}{dx} = \frac{e^{xy}y}{e^{xy}x - 2}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
Understanding the chain rule is pivotal when differentiating complex functions. It's a method for finding the derivative of composite functions, that is, functions made up by combining other functions. To conceptualize this, consider a scenario where you have functions nested within each other like layers of an onion. Applying the chain rule involves peeling back these layers one at a time and differentiating each layer as you go.

When you come across an expression like \(e^{xy}\), it's not immediately obvious how to differentiate it because it's a composition of two functions: the exponential function \(e^u\), where \(u\) is another function - in this case, \(u=xy\), a product of \(x\) and \(y\). The chain rule guides us to first differentiate the outer layer, leaving the inner layer untouched, then multiply by the derivative of the inner layer:
  • Outer layer: \(e^u\) differentiates to \(e^u\).
  • Inner layer: \(xy\) differentiates to \(x\frac{dy}{dx} + y\) when using the product rule.
Combining these steps, the derivative of \(e^{xy}\) with respect to \(x\) is \(e^{xy}(x\frac{dy}{dx} + y)\). Hence, the chain rule enables us to tackle differentiation step-by-step when faced with nested functions.
Product Rule
Differentiating products of functions often comes up in calculus, and this is where the product rule comes into play. It states that if you have a product of two functions, say \(u(x)v(x)\), the derivative of this product is not simply the product of the derivatives. Instead, you must apply the product rule, which can be stated as \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

In the given exercise, we needed to differentiate \(xy\), which is a product of \(x\) and \(y\). Applying the product rule:
  • First, hold \(x\) constant and differentiate \(y\), which gives us \(x\frac{dy}{dx}\).
  • Second, hold \(y\) constant and differentiate \(x\), which simply gives us \(y\) since the derivative of \(x\) with respect to itself is 1.

Visualizing the Product Rule

Imagine you're painting two sides of a corner where each side represents a function. The total area painted (the product of the functions) changes not just with the length of each side but also with how each side grows—in other words, their derivatives. You're totaling the growth of one side times the constant length of the other and vice versa. This visualization can help grasp why the derivatives are summed.
Differentiating Exponential Functions
Exponential functions, specifically those with a constant base like \(e\), exhibit a unique property: they are their own derivative. This means if you differentiate \(e^x\), you get \(e^x\) back. However, when you have an exponential function with a variable exponent, such as \(e^{xy}\), additional steps are necessary.

Here's where the chain rule plays its part. You treat the variable exponent \(xy\) as a separate function that also needs to be differentiated. In the exercise, we saw that after using the chain rule, terms involving \(\frac{dy}{dx}\) appear. These terms represent how the variable exponent \(xy\) changes with respect to \(x\), and thus, they must be solved to find the overall derivative of the exponential function with a variable exponent.

Applications of Exponential Functions

Exponential functions emerge across various scientific fields, from modeling population growth to calculating compound interest. Consequently, mastering differentiation of exponential functions is not just an academic exercise but a practical skill that extends into real-world problem-solving where rates of change are key.

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Most popular questions from this chapter

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$

Suppose you own a fuel-efficient hybrid automobile with a monitor on the dashboard that displays the mileage and gas consumption. The number of miles you can drive with \(g\) gallons of gas remaining in the tank on a particular stretch of highway is given by \(m(g)=50 g-25.8 g^{2}+12.5 g^{3}-1.6 g^{4},\) for \(0 \leq g \leq 4\). a. Graph and interpret the mileage function. b. Graph and interpret the gas mileage \(m(g) / \mathrm{g}\). c. Graph and interpret \(d m / d g\).

a. Derive a formula for the second derivative, \(\frac{d^{2}}{d x^{2}}(f(g(x))).\) b. Use the formula in part (a) to calculate \(\frac{d^{2}}{d x^{2}}\left(\sin \left(3 x^{4}+5 x^{2}+2\right)\right)\).

A thin copper rod, 4 meters in length, is heated at its midpoint, and the ends are held at a constant temperature of \(0^{\circ} .\) When the temperature reaches equilibrium, the temperature profile is given by \(T(x)=40 x(4-x),\) where \(0 \leq x \leq 4\) is the position along the rod. The heat flux at a point on the rod equals \(-k T^{\prime}(x),\) where \(k>0\) is a constant. If the heat flux is positive at a point, heat moves in the positive \(x\) -direction at that point, and if the heat flux is negative, heat moves in the negative \(x\) -direction. a. With \(k=1,\) what is the heat flux at \(x=1 ?\) At \(x=3 ?\) b. For what values of \(x\) is the heat flux negative? Positive? c. Explain the statement that heat flows out of the rod at its ends.

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