# prove by MI that

• Aug 20th 2006, 05:44 PM
ling_c_0202
prove by MI that

First, it is given that f(x) is a differentiable function.
and
$\displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)$

Then, I have to prove by MI that
$\displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)$

• Aug 20th 2006, 05:55 PM
ThePerfectHacker
Quote:

Originally Posted by ling_c_0202

First, it is given that f(x) is a differentiable function.
and
$\displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)$

Then, I have to prove by MI that
$\displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)$

Do this like this.
Proof (by induction) that,
$\displaystyle [f_1(x)\cdot f_2(x)\cdot...\cdot f_n(x)]'$=$\displaystyle f'_1(x)f_2(x)\cdot...\cdot f_n(x)+$$\displaystyle f_1(x)\cdot f'_2(x)\cdot ...\cdot f_n(x)+$$\displaystyle f_1(x)f_2(x)\cdot ...\cdot f'_n(x)$
Then,
$\displaystyle [f(x)]^n=\underbrace{f(x)\cdot f(x)\cdot...\cdot f(x)}_{n}$
Then by the theorem before the derivative is,
$\displaystyle f'(x)f(x)\cdot ...\cdot f(x)+f(x)f'(x)\cdot...\cdot f(x)+$$\displaystyle f(x)\cdot f(x)\cdot...\cdot f'(x) - n times. But each summand can be reduced to, \displaystyle [f(x)]^{n-1}f'(x) n-times. Thus, \displaystyle n[f(x)]^{n-1}f'(x) --- Keep you Calculus question in Calculus section :mad: -=USER WARNED=- • Aug 20th 2006, 09:19 PM CaptainBlack Quote: Originally Posted by ling_c_0202 please help with this question.. First, it is given that f(x) is a differentiable function. and \displaystyle \frac{d}{dx}(f(x))^2 = 2 f(x) f '(x) The above is the base case for the induction with \displaystyle n=2. Quote: Then, I have to prove by MI that \displaystyle \frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x) Suppose this is true for some \displaystyle k, then \displaystyle \frac{d}{dx}(f(x))^k = k (f(x)) ^{k-1}f '(x) Now (using the product rule): \displaystyle \frac{d}{dx}(f(x))^{k+1} = \frac{d}{dx} f(x) (f(x))^k$$\displaystyle = f(x)\frac{d}{dx}(f(x))^k + \frac{df}{dx} (f(x))^k$

So applying the supposition to the first term:

$\displaystyle \frac{d}{dx}(f(x))^{k+1} $$\displaystyle = k f(x) (f(x))^{k-1}f'(x) + f'(x) (f(x))^k$$\displaystyle =(k+1) (f(x))^k f'(x)$

Which is the required result for $\displaystyle k+1$.

So if we assume the required result for some $\displaystyle k$, it is true for $\displaystyle k+1$, hence
with the base case the result is proven by Mathematical Induction for all $\displaystyle n \ge 2$
(note we could have taken $\displaystyle n=1$ as the base case - which is trivialy true -
then the proof would apply for all $\displaystyle n \ge 1$).

RonL
• Aug 21st 2006, 07:55 AM
ling_c_0202
thank you
Thank you very much!
I found that i got a mistake when i apply the product rule... stupid me :p