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Chapter 3: Problem 14

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

Short Answer

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

Step by step solution

01

Identify the outer and inner functions

In this problem, we have a composite function \(y = e^{\sqrt{x}}\). The outer function \(u(v)\) is \(e^v\), and the inner function \(g(x)\) is \(\sqrt{x}\).
02

Differentiate the outer function with respect to the inner function

Now, we need to take the derivative of \(u(v) = e^v\) with respect to \(v\). Using the derivative formula for the exponential function, we get: $$\frac{d u}{d v} = e^v$$
03

Differentiate the inner function with respect to x

Next, we need to take the derivative of \(g(x) = \sqrt{x}\) with respect to \(x\). To do this, we first rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\), and then apply the power rule: $$\frac{d g}{d x} = \frac{1}{2}x^{-\frac{1}{2}}$$
04

Apply the Chain Rule

Now, we can apply the Chain Rule by multiplying the derivatives from Steps 2 and 3 together: $$\frac{dy}{dx} = \frac{d u}{d v} \cdot \frac{d g}{d x} = e^v \cdot \frac{1}{2}x^{-\frac{1}{2}}$$ Remember, \(v = g(x) = \sqrt{x}\), so we need to rewrite \(e^v\) in terms of \(x\): $$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2}x^{-\frac{1}{2}}$$ Hence, the derivative of \(y = e^{\sqrt{x}}\) with respect to \(x\) is: $$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2}x^{-\frac{1}{2}}$$

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