True of False Derivative Questions?

Question

If a function is periodic, with period c, then so is it's derivative.

I know this statement is true, but how do I explain this?? Can anyone help me?

Next question:
If the function f(x)/g(x) is defined but not differentiable at x = 1, then either f(x) or g(x) is not differentiable at x = 1.
Is this statement true or false?? Why??

Answer 1

Here's one way to see it:

f'(x+c) = lim(h->0) [f(x+c+h)-f(x+c)]/h
= lim(h->0) [f(x+h)-f(x)]/h
= f'(x).

The other statement is true, because
(f(x)/g(x))' = [f(x)g'(x)-g(x)f'(x)]/[g(x)]^2.

Since f(x)/g(x) is defined at x=1, then g(1) must be nonzero, so the derivative can't fail to be defined due to a zero denominator. The only other way for the derivative to not be defined is for either of the derivatives in the numerator to not be defined.

Answer 2

Let f be periodic with period c.
By definition, that means f(x + c) = f(x) for all x.
The derivative is defined as
f'(x) = lim {f(x + h) - f(x)} / h for h --> 0.
Now
f'(x + c) = lim {f(x + c + h) - f(x + c)} / h
... = lim {f(x + h) - f(x)} / h = f'(x)

---------------------------------

Assume f(x) and g(x) are differentiable at x = 1.
We (try to) calculate the derivative of f(x) / g(x):

lim {f(x + h)/g(x + h) - f(x)/g(x)} / h
... = lim {f(x + h) g(x) - f(x) g(x+h)} / {g(x) g(x + h) h}
... = lim { [f(x + h) - f(x)] g(x) - [g(x+h) - g(x)] f(x) } / {g(x) g(x + h) h}

This limit is well-defined unless g(x) = 0. If g(x) = 0, then f(x) / g(x) is not defined. Otherwise, we can assume that h becomes small enough so that g(x + h) <> 0 (because h is differentiable it is also continuous).

... = lim { f'(x) g(x) - g'(x) f(x) } / { g(x) g(x + h) }
... = { f'(x) g(x) - g'(x) f(x) } / { g(x) g(x) }

which is the famous quotient rule.

Answer 3

First question:

f(x) = f(x+c) for all x.
Thus, [f(x+h)-f(x)/h] = [f(x+c+h)-f(x+c)/h], for all x.
Take the limit as h ->0 of both sides, and
f'(x) = f'(x+c), or
the derivative of f is periodic with period c.

The second question:
Let h(x) = f(x)/g(x)
f(x)/g(x) is defined at x=1, so g(x) must be defined (and non-zero)at x=1.
h(x) is not differentiable at x.
We proceed with proof by contradiction:
Suppose both f and g are differentable at x=1. (thus, f'(x) and g'(x) are defined at x=1)
h'(x) = (f'(x)g(x) - g'(x)f(x))/(g(x)^2), or
h'(x) = f'(x)/g(x) - (g'(x)/g(x))*(f(x)/g(x))
Each of these terms are defined at x=1, so h'(x) is differentiable at x=1.
Contradiction.
Therefore, both f and g cannot be differentable at x=1.

Answer 4

d(e^At)/dt = Ae^At
Note that the exponent of e remains unchanged by differentiation.
Let A = j2π/c, where j=sqrt(-1) and c = period
e^j2πt/c = cos(2πt/c) - jsin(2πt/c)
or, if you prefer,
cos(2πt/c) = (e^j(2πt/c) + e^-j(2πt/c))/2.
Any periodic function can be expressed as the sum of a series of sinusoidal functions. (Fourier series)

d(f(x)/g(x))/dt = f(x)8g'(x)-g(x)f'(x))/(g(x)^2
If either is not differentiable, the quotient cannot have a derivative. The statement holds for all x.

Answer 5

if a function is periodic then...

f(x) = f(x+c) for some c. (The period)

Then...

f ' (x) = f ' (x+c)

Or... since derivative is the slope of the curve at any point, and a periodic graph has a slope which is also periodic, then the derviative must also be periodic.

Hope it helps!

Answer 6

un able to determine. we know x is 1 so 1/1=1. but we needto know what either g(x) or f(x) is to solve for the rest

Answer 7

i think im dizzy from trying to read this