Questions: y=3 e^x/(8 x^2+8) y^prime

Solution Steps

To find the derivative \( y' \) of the given function \( y = \frac{3 e^x}{8x^2 + 8} \), we will use the quotient rule. The quotient rule states that if you have a function \( y = \frac{u(x)}{v(x)} \), then its derivative is given by \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = 3e^x \) and \( v(x) = 8x^2 + 8 \).

1. Identify \( u(x) \) and \( v(x) \).
2. Compute \( u'(x) \) and \( v'(x) \).
3. Apply the quotient rule to find \( y' \).

We define the functions: \[ u(x) = 3e^x \] \[ v(x) = 8x^2 + 8 \]

Next, we calculate the derivatives of \( u(x) \) and \( v(x) \): \[ u'(x) = 3e^x \] \[ v'(x) = 16x \]

Using the quotient rule, we find the derivative \( y' \): \[ y' = \frac{u'v - uv'}{v^2} \] Substituting the values we computed: \[ y' = \frac{(3e^x)(8x^2 + 8) - (3e^x)(16x)}{(8x^2 + 8)^2} \] This simplifies to: \[ y' = \frac{-48xe^x + 3(8x^2 + 8)e^x}{(8x^2 + 8)^2} \]

Further simplification yields: \[ y' = \frac{3(e^x)(x^2 - 2x + 1)}{8(x^2 + 1)^2} \]

Final Answer