1. prove by MI that

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

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

2. Originally Posted by ling_c_0202

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

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

Do this like this.
Proof (by induction) that,
$[f_1(x)\cdot f_2(x)\cdot...\cdot f_n(x)]'$= $f'_1(x)f_2(x)\cdot...\cdot f_n(x)+$ $f_1(x)\cdot f'_2(x)\cdot ...\cdot f_n(x)+$ $f_1(x)f_2(x)\cdot ...\cdot f'_n(x)$
Then,
$[f(x)]^n=\underbrace{f(x)\cdot f(x)\cdot...\cdot f(x)}_{n}$
Then by the theorem before the derivative is,
$f'(x)f(x)\cdot ...\cdot f(x)+f(x)f'(x)\cdot...\cdot f(x)+$ $f(x)\cdot f(x)\cdot...\cdot f'(x)$ - n times. But each summand can be reduced to, $[f(x)]^{n-1}f'(x)$ n-times. Thus,
$n[f(x)]^{n-1}f'(x)$
---
Keep you Calculus question in Calculus section
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3. Originally Posted by ling_c_0202

First, it is given that f(x) is a differentiable function.
and
$
\frac{d}{dx}(f(x))^2 = 2 f(x) f '(x)
$
The above is the base case for the induction with $n=2$.

Then, I have to prove by MI that
$
\frac{d}{dx}(f(x))^n = n (f(x)) ^{n-1}f '(x)
$
Suppose this is true for some $k$, then

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

Now (using the product rule):

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

So applying the supposition to the first term:

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

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

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

RonL

4. thank you

Thank you very much!
I found that i got a mistake when i apply the product rule... stupid me