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About LinkBacksplease help with this question..
First, it is given that f(x) is a differentiable function.
and
Then, I have to prove by MI that
I don' t understand how to prove it. please help~
Do this like this.Originally Posted by ling_c_0202
please help with this question..
First, it is given that f(x) is a differentiable function.
and
Then, I have to prove by MI that
I don' t understand how to prove it
Proof (by induction) that,
=
f_2(x)\cdot...\cdot f_n(x)+)
\cdot f'_2(x)\cdot ...\cdot f_n(x)+)
Then,
Then by the theorem before the derivative is,
f(x)\cdot ...\cdot f(x)+f(x)f'(x)\cdot...\cdot f(x)+)
- n times. But each summand can be reduced to,
n-times. Thus,
---
Keep you Calculus question in Calculus section
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The above is the base case for the induction withplease help with this question..
First, it is given that f(x) is a differentiable function.
and
.
Suppose this is true for someThen, I have to prove by MI that
)^n = n (f(x)) ^{n-1}f '(x)<br />
)
, then
Now (using the product rule):
)^{k+1} = \frac{d}{dx} f(x) (f(x))^k )
So applying the supposition to the first term:
)^{k+1} )
 (f(x))^{k-1}f'(x) + f'(x) (f(x))^k<br />
)
Which is the required result for.
So if we assume the required result for some
with the base case the result is proven by Mathematical Induction for all
(note we could have takenas the base case - which is trivialy true -
then the proof would apply for all).
RonL
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