Questions: f(x)=e^x/(2 x^2+5) f'(x)=

Transcript text: $f(x)=\frac{e^{x}}{2 x^{2}+5} \\ f^{\prime}(x)=\square$

Solution

Solution Steps

To find the derivative $ f'(x) $ of the function $ f(x) = \frac{e^x}{2x^2 + 5} $, we will use the quotient rule. The quotient rule states that if you have a function $ \frac{u(x)}{v(x)} $, its derivative is given by $ \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Here, $ u(x) = e^x $ and $ v(x) = 2x^2 + 5 $.

Step 1: Identify the Functions and Their Derivatives

Given the function $ f(x) = \frac{e^x}{2x^2 + 5} $, we identify:

$ u(x) = e^x $ with derivative $ u'(x) = e^x $
$ v(x) = 2x^2 + 5 $ with derivative $ v'(x) = 4x $

Step 2: Apply the Quotient Rule

The quotient rule for derivatives is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

Substituting the identified functions and their derivatives: \[ f'(x) = \frac{e^x(2x^2 + 5) - e^x(4x)}{(2x^2 + 5)^2} \]

Step 3: Simplify the Expression

Simplify the expression in the numerator: \[ f'(x) = \frac{e^x(2x^2 + 5 - 4x)}{(2x^2 + 5)^2} \]

Final Answer

The derivative of the function is: \[ \boxed{f'(x) = \frac{e^x(2x^2 + 5 - 4x)}{(2x^2 + 5)^2}} \]

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