Question

find the first and second derivatives of the given function.
f(x)=e^{x}(4 + 3x)^{2}
calculate f(x).
f(x)=e^{x}(4 + 3x)^{2}+8e^{x}(4 + 3x)
calculate f(x).
f(x)=\square

Explanation:

Step1: Recall the product rule

The product rule is \((uv)^\prime = u^\prime v+uv^\prime\). Let \(u = e^{x}\) and \(v=(4 + 3x)^{2}+8(4 + 3x)\). Then \(u^\prime=e^{x}\)

Step2: Differentiate \(v\)

First, expand \(v=(4 + 3x)^{2}+8(4 + 3x)=16 + 24x+9x^{2}+32 + 24x=9x^{2}+48x + 48\)
\(v^\prime=18x + 48\)

Step3: Apply the product rule

\(f^{\prime\prime}(x)=e^{x}[(4 + 3x)^{2}+8(4 + 3x)]+e^{x}(18x + 48)\)
Factor out \(e^{x}\):
\(f^{\prime\prime}(x)=e^{x}[(4 + 3x)^{2}+8(4 + 3x)+18x + 48]\)
Expand \((4 + 3x)^{2}=16+24x + 9x^{2}\)
\(f^{\prime\prime}(x)=e^{x}(16+24x + 9x^{2}+32 + 24x+18x + 48)\)
Combine like - terms:
\(f^{\prime\prime}(x)=e^{x}(9x^{2}+66x + 96)\)
Factor out 3:
\(f^{\prime\prime}(x)=3e^{x}(3x^{2}+22x + 32)\)

Answer:

\(f^{\prime\prime}(x)=3e^{x}(3x^{2}+22x + 32)\)