Prove the Leibniz rule for
f
(
n
)
(
x
)
,
where
f
(
n
)
is the
n
th derivative of
f
;
that is, show that
(
f
g
)
(
n
)
(
x
)
=
n
∑
k
=
0

(
n
k
)
f
(
k
)
(
x
)
g
(
n
−
k
)
(
x
)
.

Check Explanation

Step-by-step explanation:

We want to prove that

(fg)ⁿ(x) = Σ ⁿCₖ fᵏ(x) gⁿ⁻ᵏ(x) (summing from k=0 to k = n)

Note that

(fg)ⁿ is the nth derivative of the product of f(x) × g(x)

ⁿCₖ is n combination k

fᵏ(x) is the kth derivative of f(x)

gⁿ⁻ᵏ(x) is the (n-k)th derivative of g(x)

To prove,

Let f(x) = u and g(x) = v

And f'(x) = u', g'(x) = v', etc.

We start with when n = 1

(uv)'

Using the product rule for derivatives,

(uv)' = (u'v + uv')

For n = 2

(uv)" = [(uv)']' = (u'v + uv')' (we use the product rule on each combination and the addition rule to sum up all the derivatives)

(u'v + uv')' = u"v + u'v' + u'v' + uv"

= u"v + 2u'v' + uv"

n = 3

(uv)''' = [(uv)"]' = [u"v + 2u'v' + uv"]'

= u'''v + u"v' + 2u"v' + 2u'v" + u'v" + uv'''

= u'''v + 3u"v' + 3u'v" + uv'''

n = 4

(uv)"" = [(uv)''']' = [u'''v + 3u"v' + 3u'v" + uv''']'

= u""v + u'''v' + 3u'''v' + 3u"v" + 3u'v''' + u'v''' + uv""

= u""v + 4u'''v' + 6u"v" + 4u'v''' + uv"" + uv""

At tbis stage, it is evident that the coefficients of each term in the final expression for the nth derivative of the product of two functions matches the Pascal's triangle equivalent.

Furthermore, it is clear that these formulas are similar to the binomial expansion raised to the appropriate exponent. With the terms with zero exponent u⁰ and v⁰ corresponding to the functions u and v themselves.

We can then write the general formula for the nth derivative of the product of two functions, uv, using binomial theorem's general formula as

(uv)ⁿ = Σ ⁿCₖ uᵏ vⁿ⁻ᵏ (summing from k = 0 to k = n)

(uv)ⁿ isn't the nth power of (uv) like the normal binomial expansion, instead, it is the nth derivative of the product of u and v.

ⁿCₖ is still n combination k

uᵏ isn't the kth power of u like in the normal binomial expansion, instead, it is the kth derivative of u

vⁿ⁻ᵏ isn't the (n-k)th power of v, instead, it is the (n-k)th derivative of v

We can then substitute back u = f(x) and v = g(x)

(fg)ⁿ(x) = Σ ⁿCₖ fᵏ(x) gⁿ⁻ᵏ(x) (summing from k=0 to k = n)

(Proved!)

Hope this Helps!!!

answer: no idea

step-by-step explanation: i just want points

just needed some points but try socratic