Respuesta :
Answer:
[tex]f'(x) = e^{x}(x^{3} + 3x^{2} + 1)[/tex]
and [tex]f''(x) = e^{x}(x^{3}+6x^{2} + 6x + 1)[/tex]
Step-by-step explanation:
From the question,
f(x) = (x3 + 1)ex
That is,
[tex]f(x) = (x^{3} +1)e^{x}[/tex]
Then , we can write that
[tex]f(x) = x^{3}e^{x} +e^{x}[/tex]
To find f'(x), we will differentiate [tex]x^{3}e^{x}[/tex] and then add it to the differential of [tex]e^{x}[/tex]
First, we will differentiate [tex]x^{3}e^{x}[/tex],
Let [tex]y(x) = x^{3}e^{x}[/tex] (This is a product)
Then,
[tex]f(x) = y(x) + e^{x}[/tex]
Hence,
[tex]f'(x) = y'(x) + (e^{x})'[/tex]
Given [tex]y(x) = u(x) v(x)[/tex], Using the product rule
[tex]y'(x) = u(x).v'(x) + u'(x).v(x)[/tex]
Hence,
[tex]y'(x) = x^{3}(e^{x})' + (x^{3})'e^{x}[/tex]
[tex]y'(x) = x^{3}e^{x} + 3x^{2}e^{x}[/tex]
and
[tex](e^{x})' = e^{x}[/tex]
[tex]f'(x) = y'(x) + (e^{x})'[/tex]
∴ [tex]f'(x) = x^{3}e^{x} + 3x^{2}e^{x} + e^{x}[/tex]
[tex]f'(x) = e^{x}(x^{3} + 3x^{2} + 1)[/tex]
For f ''(x)
To find f ''(x), will differentiate f '(x)
[tex]f'(x) = e^{x}(x^{3} + 3x^{2} + 1)[/tex]
This is also a product, then we will apply the product rule
From the product rule,
Given [tex]y(x) = u(x) v(x)[/tex], Using the product rule
[tex]y'(x) = u(x).v'(x) + u'(x).v(x)[/tex]
Here, Let[tex]u(x) = e^{x}[/tex] and [tex]v(x) = x^{3} + 3x^{2} + 1[/tex]
Then,
[tex]f''(x) = e^{x}(x^{3} + 3x^{2} + 1)' + (e^{x})'(x^{3} + 3x^{2} + 1)[/tex]
[tex]f''(x) = e^{x}(3x^{2} + 6x ) + (e^{x})(x^{3} + 3x^{2} + 1)[/tex]
∴[tex]f''(x) = e^{x}(x^{3}+6x^{2} + 6x + 1)[/tex]
Hence,
[tex]f'(x) = e^{x}(x^{3} + 3x^{2} + 1)[/tex]
and [tex]f''(x) = e^{x}(x^{3}+6x^{2} + 6x + 1)[/tex]
