Question

Use implicit differentiation to find$\frac{dy}{dx}.$ $$x+2y=\sqrt{y}$$

Short answer

Question: Given the equation $x+2y=\sqrt{y}$, find the first derivative of y with respect to x using implicit differentiation. Answer: $\frac{dy}{dx}=\frac{-1}{2-\frac{1}{2\sqrt{y}}}$

Step-by-step solution

Differentiate both sides of the equation

We will differentiate both sides of $x+2y=\sqrt{y}$ with respect to x. Since y is also a function of x, we need to use the chain rule while differentiating y terms. Using chain rule, we can rewrite the differentiation of y as $\frac{d}{dx}(y)=\frac{dy}{dx}$. So, we have: $$\frac{d}{dx}(x)+\frac{d}{dx}(2y)=\frac{d}{dx}(\sqrt{y})$$

Apply differentiation

Now, we apply the differentiation rules to each term, remembering to include the $\frac{dy}{dx}$ term for the y-variable: $$1+(2\cdot \frac{dy}{dx})=\frac{1}{2\sqrt{y}}\cdot \frac{dy}{dx}$$

Solve for $\frac{dy}{dx}$

To find $\frac{dy}{dx}$, we will isolate the term by moving all terms containing $\frac{dy}{dx}$ to one side of the equation and the remaining terms to the other side. $$2\frac{dy}{dx}-\frac{1}{2\sqrt{y}}\frac{dy}{dx}=-1$$ Now, factor out $\frac{dy}{dx}$ from the left side of the equation: $$\frac{dy}{dx}(2-\frac{1}{2\sqrt{y}})=-1$$ Finally, divide both sides of the equation by the expression in the parenthesis to solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{-1}{2-\frac{1}{2\sqrt{y}}}$$ So, our resulting expression for $\frac{dy}{dx}$ is: $$\frac{dy}{dx}=\frac{-1}{2-\frac{1}{2\sqrt{y}}}$$

Key concepts

Differentiation

Differentiation is the process of finding the derivative of a function. In simpler terms, it's all about measuring how a function changes when its input changes. Consider it as finding the slope at any given point of the function curve.
The goal of differentiation is to determine the function's rate of change. In our example, the equation is:

• $x+2y=\sqrt{y}$

When differentiating, we must apply the rules of calculus to every variable present, particularly if they are interacting with each other. When given an equation with multiple terms, differentiation involves treating each term separately, then combining the results.
It's important to note that if a term involves the variable $y$, which is implicitly dependent on $x$, then the differentiation process is a bit nuanced, requiring a deeper understanding of implicit functions. This is where implicit differentiation comes into play.

Chain Rule

The chain rule is a fundamental differentiation technique used when we deal with compositions of functions. Simply put, it helps us differentiate a function inside another function.
Consider the term $\sqrt{y}$ in our equation. Here, "y" is a function in itself, so the chain rule is used to differentiate $\sqrt{y}$ with respect to $x$.

• Firstly, differentiate the outer function, which is the square root function. The derivative of $\sqrt{y}$ with respect to $y$ is $\frac{1}{2\sqrt{y}}$.
• Then, differentiate the inner function, which is $y$ with respect to $x$, represented as $\frac{dy}{dx}$.

By multiplying these derivatives together, the chain rule provides the derivative of a composite function with respect to $x$.
This rule ensures that we correctly account for the interaction between functions within our equation, emphasizing accuracy in our differentiation process.

Derivatives

Derivatives are central to the concept of differentiation. They represent the instantaneous rate of change of a function as one of its variables changes.
In the context of implicit differentiation, as seen in our example equation, derivatives tell us how $y$ changes with respect to $x$. Here, we aim to find $\frac{dy}{dx}$:

• First, differentiate each term of the equation separately, identifying how each component contributes to the change.For $x$, the derivative is constant, so $\frac{d}{dx}x=1$.
• For $2y$, applying the linearity of differentiation, it becomes $2\frac{dy}{dx}$, considering $y$ as a function of $x$.
• Finally, applying the derivative to the square root part, as we did with chain rule:$\frac{d}{dx}(\sqrt{y})=\frac{1}{2\sqrt{y}}\frac{dy}{dx}$.

Derivatives reveal the relationship between the variables as they change, making it easier to isolate and solve for $\frac{dy}{dx}$. They've broad applications in physics, engineering, and beyond, wherever change is modeled mathematically.