Question

attempt 1: 10 attempts remaining. find the derivative of the function ( y = e^{sqrt{x^2 + 3}} ) using the chain rule for exponential functions. ( \frac{dy}{dx} = ) submit answer next item

Explanation:

Step1: Identify the outer and inner functions

The function is \( y = e^{\sqrt{x^2 + 3}} \). Let the outer function be \( f(u)=e^u \) and the inner function be \( u = g(x)=\sqrt{x^2 + 3}=(x^2 + 3)^{\frac{1}{2}} \).

Step2: Differentiate the outer function

The derivative of \( f(u)=e^u \) with respect to \( u \) is \( f'(u)=e^u \).

Step3: Differentiate the inner function

Using the power rule and chain rule for \( u=(x^2 + 3)^{\frac{1}{2}} \), the derivative \( g'(x) \) is calculated as follows:
First, apply the power rule: if \( y = v^n \), then \( y'=n v^{n - 1}v' \). Here, \( v=x^2 + 3 \), \( n=\frac{1}{2} \).
So, \( g'(x)=\frac{1}{2}(x^2 + 3)^{-\frac{1}{2}}\times(2x) \).
Simplify this: \( g'(x)=\frac{2x}{2\sqrt{x^2 + 3}}=\frac{x}{\sqrt{x^2 + 3}} \).

Step4: Apply the chain rule

The chain rule states that \( \frac{dy}{dx}=f'(u)\times g'(x) \). Substitute \( u = \sqrt{x^2 + 3} \), \( f'(u)=e^u \) and \( g'(x)=\frac{x}{\sqrt{x^2 + 3}} \) into the formula.
We get \( \frac{dy}{dx}=e^{\sqrt{x^2 + 3}}\times\frac{x}{\sqrt{x^2 + 3}} \).

Answer:

\( \frac{x e^{\sqrt{x^2 + 3}}}{\sqrt{x^2 + 3}} \)