Problem 14
Question
Use implicit differentiation to find \(d y / d x\). $$x \cos (2 x+3 y)=y \sin x$$
Step-by-Step Solution
Verified Answer
The derivative \(\frac{dy}{dx}\) is given by \(\frac{dy}{dx} = \frac{y \cos x + 2x \sin(2x+3y) - \cos(2x+3y)}{-3x \sin(2x+3y) - \sin x}\).
1Step 1: Differentiate both sides with respect to x
Start by differentiating both sides of the equation with respect to \(x\). Using implicit differentiation, remember to apply the product rule and chain rule. The equation is \(x \cos (2x + 3y) = y \sin x\). Differentiate each side with respect to \(x\).
2Step 2: Differentiate the left side
The left side is \(x \cos (2x + 3y)\). Applying the product rule, differentiate as follows: \(\frac{d}{dx}[x] \cdot \cos(2x + 3y) + x \cdot \frac{d}{dx}[\cos(2x + 3y)]\). So, it simplifies into:\[1 \cdot \cos(2x+3y) - x \cdot \sin(2x + 3y) \cdot (2 + 3 \frac{dy}{dx})\]
3Step 3: Differentiate the right side
Now take the derivative of the right side, \(y \sin x\), using the product rule:\(\frac{d}{dx}[y] \cdot \sin x + y \cdot \frac{d}{dx}[\sin x]\). This simplifies into:\[\frac{dy}{dx} \sin x + y \cos x\]
4Step 4: Combine and rearrange the equation
Combine the expressions from Step 2 and Step 3:\[\cos(2x+3y) - x \sin(2x+3y)(2 + 3\frac{dy}{dx}) = \frac{dy}{dx} \sin x + y \cos x\]Rearrange all terms involving \(\frac{dy}{dx}\) to one side and constant terms to the other side.
5Step 5: Solve for \(\frac{dy}{dx}\)
Factor out \(\frac{dy}{dx}\) from the terms you moved to one side:\[ -x \sin(2x + 3y) \cdot 3\frac{dy}{dx} - \frac{dy}{dx} \sin x = y \cos x + x \sin (2x+3y) \cdot 2 - \cos(2x+3y)\]Factor out \(\frac{dy}{dx}\) to get:\[\frac{dy}{dx}( -3x \sin(2x + 3y) - \sin x) = y \cos x + 2x \sin(2x+3y) - \cos(2x+3y)\]Solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{y \cos x + 2x \sin(2x+3y) - \cos(2x+3y)}{-3x \sin(2x+3y) - \sin x}\]
Key Concepts
Product RuleChain RuleDerivatives
Product Rule
When dealing with derivatives in calculus, the product rule is an essential tool. It's used when you're differentiating a product of two functions. The rule states: if you have two functions, say \(u(x)\) and \(v(x)\), then the derivative of their product is given by:
This means you take the derivative of the first function, multiply it by the second function as it is, and then add the first function as it is times the derivative of the second function.
In the context of the given problem, where the left side is \(x \cos(2x + 3y)\), the product rule helps us separate the differentiation into parts. First, you'll take the derivative of \(x\), leaving \(\cos(2x + 3y)\) untouched. Then, you'll take \(x\) as it is and differentiate \(\cos(2x + 3y)\) using the chain rule, which we'll cover in the next section.
- \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)
This means you take the derivative of the first function, multiply it by the second function as it is, and then add the first function as it is times the derivative of the second function.
In the context of the given problem, where the left side is \(x \cos(2x + 3y)\), the product rule helps us separate the differentiation into parts. First, you'll take the derivative of \(x\), leaving \(\cos(2x + 3y)\) untouched. Then, you'll take \(x\) as it is and differentiate \(\cos(2x + 3y)\) using the chain rule, which we'll cover in the next section.
Chain Rule
The chain rule is another fundamental tool in calculus, crucial for differentiating composite functions. Essentially, it allows you to work with functions inside of other functions. The chain rule says: if you have two functions, say \(u(x)\), substituted into another function \(f(u)\), then the derivative is:
For the original problem, this becomes relevant when differentiating parts like \(\cos(2x + 3y)\). You first differentiate the outer function \(\cos(u)\) which becomes \(-\sin(u)\).
Then, multiply this by the derivative of the inner function \(2x + 3y\), which requires differentiating both \(2x\) and \(3y\), remembering that \(y\) is a function of \(x\) so \(\frac{dy}{dx}\) comes into play. The chain rule is crucial for properly working through implicit differentiation because many functions are composed of others.
- \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \)
For the original problem, this becomes relevant when differentiating parts like \(\cos(2x + 3y)\). You first differentiate the outer function \(\cos(u)\) which becomes \(-\sin(u)\).
Then, multiply this by the derivative of the inner function \(2x + 3y\), which requires differentiating both \(2x\) and \(3y\), remembering that \(y\) is a function of \(x\) so \(\frac{dy}{dx}\) comes into play. The chain rule is crucial for properly working through implicit differentiation because many functions are composed of others.
Derivatives
Derivatives measure how a function changes as its input changes. In calculus, the derivative is a key concept that allows us to understand rates of change, slopes of curves, and is the underpinning of much of calculus. For example, the derivative of \(x\) with respect to \(x\) is \(1\), and the derivative of \(\sin(x)\) with respect to \(x\) is \(\cos(x)\).
In the practice of implicit differentiation, derivatives become even more interesting. When given an equation that defines \(y\) implicitly as a function of \(x\), such as in our case with \(x \cos(2x + 3y) = y \sin x\), we take derivatives of both sides with respect to \(x\). This often involves the derivative of \(y\) as part of the computation, which is why \(\frac{dy}{dx}\) appears.
This process allows us to solve for \(\frac{dy}{dx}\), the rate at which \(y\) changes with respect to \(x\), even when \(y\) is not explicitly defined in terms of \(x\). Understanding how to navigate these derivatives through implicit differentiation is crucial for solving problems like the one in this exercise.
In the practice of implicit differentiation, derivatives become even more interesting. When given an equation that defines \(y\) implicitly as a function of \(x\), such as in our case with \(x \cos(2x + 3y) = y \sin x\), we take derivatives of both sides with respect to \(x\). This often involves the derivative of \(y\) as part of the computation, which is why \(\frac{dy}{dx}\) appears.
This process allows us to solve for \(\frac{dy}{dx}\), the rate at which \(y\) changes with respect to \(x\), even when \(y\) is not explicitly defined in terms of \(x\). Understanding how to navigate these derivatives through implicit differentiation is crucial for solving problems like the one in this exercise.
Other exercises in this chapter
Problem 14
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(t^{3 / 2}\right)+\sqrt{t}$$
View solution Problem 14
In Exercises \(9-22,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x\). $$y=\sqrt{3 x^{2}-4 x+6}$$
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Find \(d y / d x\). $$y=\frac{\cos x}{x}+\frac{x}{\cos x}$$
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Find \(y^{\prime}\) (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. $$y=(2 x+3)\left(5 x
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