Question

find the derivative of the function y = e^{\sqrt{x^2 + 4}} using the chain rule for exponential functions. \frac{dy}{dx} =

Explanation:

Step1: Identify the outer and inner functions

Let \( u = \sqrt{x^2 + 4} \), so the outer function is \( y = e^u \) and the inner function is \( u = (x^2 + 4)^{\frac{1}{2}} \).

Step2: Differentiate the outer function with respect to \( u \)

The derivative of \( y = e^u \) with respect to \( u \) is \( \frac{dy}{du} = e^u \).

Step3: Differentiate the inner function with respect to \( x \)

Using the power rule and chain rule, for \( u = (x^2 + 4)^{\frac{1}{2}} \), we have:
\( \frac{du}{dx} = \frac{1}{2}(x^2 + 4)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2 + 4}} \)

Step4: Apply the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

Substitute \( \frac{dy}{du} = e^u \) and \( \frac{du}{dx} = \frac{x}{\sqrt{x^2 + 4}} \), and recall that \( u = \sqrt{x^2 + 4} \):
\( \frac{dy}{dx} = e^{\sqrt{x^2 + 4}} \cdot \frac{x}{\sqrt{x^2 + 4}} \)

Answer:

\( \frac{x e^{\sqrt{x^2 + 4}}}{\sqrt{x^2 + 4}} \)