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Chapter 3: Problem 14

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

Short Answer

Expert verified
Answer: The derivative of y with respect to x is $\frac{dy}{dx} = 0$.

Step by step solution

01

Differentiate both sides of the equation with respect to x

Apply \(\frac{d}{dx}\) to both sides of the equation: $$\frac{d}{dx} (e^{x y}) = \frac{d}{dx} (2 y)$$
02

Apply the chain rule for the left side and differentiate the right side

For the left side, by applying the chain rule for the term e^{xy}, we have: $$\frac{d}{dx} (e^{xy}) = (e^{xy})\left(\frac{d}{dx} (xy)\right)$$ For the right side, differentiate 2y with respect to x, treating y as a function of x: $$\frac{d}{dx} (2y) = 2\frac{dy}{dx}$$
03

Differentiate the inner function (xy) and substitute back

Differentiate xy with respect to x, treating y as a function of x, by using the product rule: $$\frac{d}{dx}(xy) = x\frac{dy}{dx}+y$$ Substitute this expression back into the left side of step 2: $$(e^{xy})\left(x\frac{dy}{dx}+y\right)=2\frac{dy}{dx}$$
04

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

Arrange the equation to solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}(xe^{xy})+\frac{dy}{dx}(ye^{xy})-2\frac{dy}{dx}=0$$ Factor out the \(\frac{dy}{dx}\) term: $$\frac{dy}{dx}(xe^{xy}+ye^{xy}-2)=0$$ Now divide by the factor (xe^{xy}+ye^{xy}-2) to solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{0}{xe^{xy}+ye^{xy}-2}$$ Thus, the derivative of y with respect to x is: $$\frac{dy}{dx}=0$$

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Key Concepts

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

Chain Rule
Implicit differentiation, especially in exercises involving exponential functions such as \( e^{x y} = 2y \), requires the adept use of the Chain Rule. The Chain Rule is essential in calculus for functions composed of other functions.

Remember, when differentiating a composite function, like \( e^{xy} \), we first take the derivative of the outer function and multiply it by the derivative of the inner function.
  • With \( e^{xy} \), the outer function is \( e^u \) (where \( u = xy \)), which differentiates to \( e^{u} \), maintaining the original expression.
  • The inner derivative, \( \frac{d}{dx} (xy) \), requires further differentiation, which leads us to the next key tool: the Product Rule.
Implicit differentiation frequently involves chaining different rules together,making it a conceptually layered process.
Product Rule
Using the Product Rule is crucial when differentiating terms such as \( xy \) within the larger context of implicit differentiation. This rule helps when a function is the product of two separate functions of the independent variable.

In the expression \( x y \), consider \( x \) and \( y \) as two such functions. The Product Rule states: if \( u(x) \) and \( v(x) \) are functions of \( x \), then \( \frac{d}{dx}(uv) = u'v + uv' \).
  • For \( xy \), \( u(x) = x \) and \( v(x) = y \).
  • The derivative \( \frac{d}{dx}(xy) \) becomes \( x \frac{dy}{dx} + y \), thus allowing us to correctly address the impact on the derivative.
This derivative fits into larger implicit solutions, pile-driving our progress.
Calculus Differentiation
Differentiation is foundational to calculus, offering a toolkit for addressing complex functions. Implicit differentiation is a technique used when it's problematic to separate the dependent variable on one side.

This method typically involves differentiating both sides with respect to \( x \) concurrently.
  • The given equation \( e^{x y} = 2y \) exemplifies a situation where \( y \) is not easily separated.
  • This calls for differentiating both sides, applying necessary rules like the Chain and Product rules.Achieving implicit differentiation leads to solving for \( \frac{dy}{dx} \), which represents the rate of change of \( y \) with respect to \( x \).
In this case, real-world applications can be seen in circumstances where relationships between variables are intertwined, yet an understanding of variation is crucial.Getting familiar with the systematic approach will significantly simplify deciphering trickier mathematical phenomena.

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Most popular questions from this chapter

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

Multiple tangent lines Complete the following steps. a. Find equations of all lines tangent to the curve at the given value of \(x\) b. Graph the tangent lines on the given graph. \(4 x^{3}=y^{2}(4-x) ; x=2\) (cissoid of Diocles)

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,2)\) (lima莽on of Pascal)

The Witch of Agnesi The graph of \(y=\frac{a^{3}}{x^{2}+a^{2}},\) where \(a\) is a constant, is called the witch of Agnesi (named after the 18th-century Italian mathematician Maria Agnesi). a. Let \(a=3\) and find an equation of the line tangent to \(y=\frac{27}{x^{2}+9}\) at \(x=2\) b. Plot the function and the tangent line found in part (a).

Proof of \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\) Use the trigonometric identity \(\cos ^{2} x+\sin ^{2} x=1\) to prove that \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0 .(\) Hint: Begin by multiplying the numerator and denominator by \(\cos x+1 .)\)
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