Question

Having trouble with this math question. I was told the answer was 30e^4problem is IF f(x) = e^2x (x^3 + 1), then f'(2) = I don't know how they got the answer.

Answer 1

f(x) = e^2x (x^3 + 1)

It can also be written as it follows:

`f(x)=e^(2x)\cdot(x^3+1)`

first we need to determine f'(x).

we can see this function as a product of two funcions:

`\begin{gathered} f(x)=g(x)\cdot h(x) \\ g(x)=e^(2x);h(x)=x^3+1 \end{gathered}`

so, we can apply the Product rule

`(d)/(dx)\lbrack g(x)\cdot h(x)\rbrack=g(x)\cdot h^(\prime)(x)+g^(\prime)(x)\cdot h(x)`

Let's determine the derivate of those two function

`\begin{gathered} g(x)=e^(2x) \\ g^(\prime)(x)=2\cdot e^(2x) \end{gathered}`
`\begin{gathered} h(x)=x^3+1 \\ h^(\prime)(x)=3x^2 \end{gathered}`

Now, we can use the functions g(x) and h(x) and its derivates to determine f'(x) using the Product rule

`\begin{gathered} f^(\prime)(x)=g(x)\cdot h^(\prime)(x)+g^(\prime)(x)\cdot h(x) \\ f^(\prime)(x)=e^(2x)\cdot3x^2+2e^(2x)\cdot(x^3+1) \end{gathered}`

we can simplify this expression as it follows:

`\begin{gathered} f^(\prime)(x)=3e^(2x)x^2+2e^(2x)x^3+2e^(2x) \\ f^(\prime)(x)=2e^{\mleft\{2x\mright\}}x^3+3e^{\mleft\{2x\mright\}}x^2+2e^{\mleft\{2x\mright\}} \end{gathered}`

Finally, we need to replace x = 2 to find f'(2)

`f^(\prime)(2)=2e^{\{2\cdot2\}}\cdot2^3+3e^{\{2\cdot2\}}\cdot2^2+2e^{\{2\cdot2\}}`

simplify:

`\begin{gathered} f^(\prime)(2)=2e^4\cdot8+3e^4\cdot4+2e^4 \\ f^(\prime)(2)=16e^4+12e^4+2e^4^{} \\ f^(\prime)(2)=(16+12+2)\cdot e^4 \\ f^(\prime)(2)=30e^4 \end{gathered}`