Respuesta :
a. By the product rule,
[tex]F=fg\implies F'=f'g+fg'[/tex]
[tex]\implies F''=(f''g+f'g')+(f'g'+fg'')=f''g+2f'g'+fg''[/tex]
b. By the same rule,
[tex]F'''=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')[/tex]
[tex]F'''=f'''g+3f''g'+3f'g''+fg'''[/tex]
and
[tex]F^{(4)}=(f^{(4)}g+f'''g')+3(f'''g'+f''g'')+3(f''g''+f'g''')+(f'g'''+fg^{(4)})[/tex]
[tex]F^{(4)}=f^{(4)}g+4f'''g'+6f''g''+4f'g'''+fg^{(4)}[/tex]
c. You might recognize the coefficients as those that appear in the expansion of [tex](a+b)^n[/tex]:
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1
and so on; the general pattern (known as the general Leibniz rule) is
[tex]F^{(n)}=\displaystyle\sum_{k=0}^n\binom nkf^{(n-k)}g^{(k)}[/tex]
a) The second derivative is [tex]F''(x) = f''(x) \cdot g(x) + 2\cdot f'(x)\cdot g'(x) + f(x) \cdot g''(x)[/tex].
b) The third and fourth derivatives are [tex]F'''(x) = f'''(x)\cdot g(x) +3\cdot f''(x) \cdot g'(x) + 3\cdot f'(x)\cdot g''(x) + f(x)\cdot g'''(x)[/tex] and [tex]F^{(4)}(x) = f^{(4)}(x) \cdot g(x) + 4\cdot f'''(x)\cdot g'(x) + 6\cdot f''(x)\cdot g''(x) + 4\cdot f'(x)\cdot g'''(x) +f(x) \cdot g^{(4)}(x)[/tex], respectively.
c) There is a power binomial pattern for higher derivatives of F.
In this question we are going to derive expressions for [tex]F(x) = f(x)\cdot g(x)[/tex] by derivative rules, to be more specific, the derivative rule for the product between two functions.
a) By the derivative rule for the product between two functions we have the following result:
First derivative
[tex]F'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)[/tex]
Second derivative
[tex]F''(x) = f''(x)\cdot g(x) + f'(x)\cdot g'(x) + f'(x)\cdot g'(x) + f(x)\cdot g''(x)[/tex]
[tex]F''(x) = f''(x) \cdot g(x) + 2\cdot f'(x)\cdot g'(x) + f(x) \cdot g''(x)[/tex] (1)
The second derivative is [tex]F''(x) = f''(x) \cdot g(x) + 2\cdot f'(x)\cdot g'(x) + f(x) \cdot g''(x)[/tex].
b) And we find the formulas for the third and fourth derivatives of the expression from (1):
Third derivative
[tex]F''(x) = f''(x) \cdot g(x) + 2\cdot f'(x)\cdot g'(x) + f(x) \cdot g''(x)[/tex]
[tex]F'''(x) = f'''(x)\cdot g(x) + f''(x) \cdot g'(x) + 2\cdot f''(x)\cdot g'(x) +2\cdot f'(x)\cdot g''(x) + f'(x)\cdot g''(x) + f(x) \cdot g'''(x)[/tex]
[tex]F'''(x) = f'''(x)\cdot g(x) +3\cdot f''(x) \cdot g'(x) + 3\cdot f'(x)\cdot g''(x) + f(x)\cdot g'''(x)[/tex] (2)
Fourth derivative
[tex]F^{(4)}(x) = f^{(4)}(x) \cdot g(x) + f'''(x) \cdot g'(x) + 3\cdot f'''(x)\cdot g'(x) +3\cdot f''(x)\cdot g''(x) + 3\cdot f''(x) \cdot g''(x) + 3\cdot f'(x) \cdot g'''(x) + f'(x)\cdot g'''(x) + f(x) \cdot g^{(4)}(x)[/tex]
[tex]F^{(4)}(x) = f^{(4)}(x) \cdot g(x) + 4\cdot f'''(x)\cdot g'(x) + 6\cdot f''(x)\cdot g''(x) + 4\cdot f'(x)\cdot g'''(x) +f(x) \cdot g^{(4)}(x)[/tex] (3)
The third and fourth derivatives are [tex]F'''(x) = f'''(x)\cdot g(x) +3\cdot f''(x) \cdot g'(x) + 3\cdot f'(x)\cdot g''(x) + f(x)\cdot g'''(x)[/tex] and [tex]F^{(4)}(x) = f^{(4)}(x) \cdot g(x) + 4\cdot f'''(x)\cdot g'(x) + 6\cdot f''(x)\cdot g''(x) + 4\cdot f'(x)\cdot g'''(x) +f(x) \cdot g^{(4)}(x)[/tex], respectively.
c) Based on (1), (2) and (3), we notice a power binomial pattern defined by the power binomial theorem. In other words, we can derive this expression by analogy:
[tex]F^{(n)} (x) = \Sigma\limits_{k = 0}^{n} \left(\begin{array}{c}n\\k\end{array} \right)\,f^{(n-k)}(x)\cdot g^{(k)}(x)[/tex] (4)
Where:
- [tex]F^{(n)}(x)[/tex] - n-th derivative of the function.
- [tex]n[/tex] - Order of the derivative.
- [tex]k[/tex] - Index of the sum term.
- [tex]f^{(n-k)}(x)[/tex] - (n-k)-th derivative of [tex]f(x)[/tex].
- [tex]g^{(k)}(x)[/tex] - k-th derivative of [tex]g(x)[/tex].
There is a power binomial pattern for higher derivatives of F.
We kindly invite to check this question on derivatives: https://brainly.com/question/14415022
