Question

attempt 1: 10 attempts remaining. find the derivative. ( y = sqrt{1 + e^{2x}} ) ( y = ) submit answer next item

Explanation:

Step1: Rewrite the function

Rewrite \( y = \sqrt{1 + e^{2x}} \) as \( y=(1 + e^{2x})^{\frac{1}{2}} \).

Step2: Apply the chain rule

The chain rule states that if \( y = u^n \) and \( u = f(x) \), then \( y'=n\cdot u^{n - 1}\cdot u' \). Here, \( n=\frac{1}{2} \) and \( u = 1+e^{2x} \).
First, find the derivative of \( u \) with respect to \( x \). The derivative of \( 1 \) is \( 0 \), and the derivative of \( e^{2x} \) with respect to \( x \) (using the chain rule again, let \( v = 2x \), so derivative of \( e^{v} \) with respect to \( v \) is \( e^{v} \) and derivative of \( v \) with respect to \( x \) is \( 2 \)) is \( 2e^{2x} \). So \( u'=2e^{2x} \).
Now, find \( y' \) using the chain rule:
\( y'=\frac{1}{2}(1 + e^{2x})^{\frac{1}{2}-1}\cdot(2e^{2x}) \)

Step3: Simplify the expression

Simplify the exponents and the constants:
\( \frac{1}{2}(1 + e^{2x})^{-\frac{1}{2}}\cdot(2e^{2x})=\frac{2e^{2x}}{2\sqrt{1 + e^{2x}}}=\frac{e^{2x}}{\sqrt{1 + e^{2x}}} \)

Answer:

\( \frac{e^{2x}}{\sqrt{1 + e^{2x}}} \)