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Chapter 3: Problem 13

Use Version I of the Chain Rule to calculate \(\frac{d y}{d x}\). $$y=\sqrt{x^{2}+1}$$

Short Answer

Expert verified
Answer: The derivative of the function \(y = \sqrt{x^2 + 1}\) with respect to x is \(\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}\).

Step by step solution

01

Identify the inner and outer functions

We have \(y = \sqrt{x^2 + 1}\). Here, the inner function (denoted as 'u') is \(x^2 + 1\), and the outer function (denoted as 'v') is the square root.
02

Differentiate the inner function

To find the derivative of the inner function \(u = x^2 + 1\), simply differentiate it with respect to x: $$\frac{du}{dx} = \frac{d(x^2 + 1)}{dx} = 2x$$
03

Differentiate the outer function

Next, we need to find the derivative of the outer function \(v = \sqrt{u}\). To do this, consider the outer function as \(v = u^{\frac{1}{2}}\). Then, differentiate it with respect to u: $$\frac{dv}{du} = \frac{1}{2}u^{-\frac{1}{2}}$$
04

Apply the Chain Rule

Now, we apply the Chain Rule. The Chain Rule states that \(\frac{d(y)}{dx} = \frac{d(v)}{du} \cdot \frac{d(u)}{dx}\). So, let us now plug in our previously derived expressions: $$\frac{dy}{dx} = \frac{1}{2}\cdot u^{-\frac{1}{2}}\cdot 2x$$
05

Replace u with the inner function

We need to rewrite the expression in terms of x, so replace u with the inner function \(x^2 + 1\). This gives us: $$\frac{dy}{dx} = \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}\cdot 2x$$
06

Simplify the expression

Finally, simplify the expression: $$\frac{dy}{dx} = \frac{2x}{2\sqrt{x^2 + 1}}$$ $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$$ Therefore, the derivative of the given function is: $$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}}$$

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