megadeth95
New member
- Joined
- Nov 25, 2011
- Messages
- 35
Hello Forum, I find this problem to be extremely difficult. Any help is appreacited!
Each of the following "derivatives" has been done incorrectly. In each case below, find a counterexample, that shows why the reasoning is incorrect. Then correct the mistakes by finding the actual derivatives of the left hand side.
For example: The statement "d/dx f(2) = f'(2x) is false because if f(x)=x^2, then we have d/dx f(2x) = d/dx [4x^2] = 8x, but on the other hand, f'(x) = 2x, so f'(2x) = 4x . In general, the correct derivative would be "d/dx f(2x) = f'(2x) * 2" by the chain rule.
(a) d/dx f(x+5) = f'(x)
(b) d/dt g'(t/3) = g'(t) / 3
Attempts
Correcting the mistakes is easy but finding a counterexample is not. For question (a), I reasoned that the counterexample would be the anti-derivative of x+5, which is (1/2)x^2 + 5x , just like in the example they use x^2 (which is the anti-derivative of 2x) From there, this is what I did:
if f(x) = (1/2)x^2 + 5x
then --> d/dx f(x+5) = d/dx [ (x^2 + 20x +75)/2 ] = (4x+40)/4
but on the other hand f'(x) = x +5
so f'(x+5) = x + 5
therefore x + 5 ≠ (is not equal to) (4x+40)/4
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Similarly, I applied the same idea for question (b)
if g(t) = (1/6)t^2
then --> d/dt g(t/3) = d/dt [ (t^2)/54 ] = (108t)/54^2
but on the other hand f'(x) = t/3
so f'(t/3) = (36t)/18^2
therefore (36t)/18^2 ≠ (is not equal to) (108t)/54^2
I think these are not the counterexamples they are asking for. How is the right way to find them? Should I use other equations or formulas?
Thank you in advance for any help you might be able to provide.
Sincerely
megadeth95
Each of the following "derivatives" has been done incorrectly. In each case below, find a counterexample, that shows why the reasoning is incorrect. Then correct the mistakes by finding the actual derivatives of the left hand side.
For example: The statement "d/dx f(2) = f'(2x) is false because if f(x)=x^2, then we have d/dx f(2x) = d/dx [4x^2] = 8x, but on the other hand, f'(x) = 2x, so f'(2x) = 4x . In general, the correct derivative would be "d/dx f(2x) = f'(2x) * 2" by the chain rule.
(a) d/dx f(x+5) = f'(x)
(b) d/dt g'(t/3) = g'(t) / 3
Attempts
Correcting the mistakes is easy but finding a counterexample is not. For question (a), I reasoned that the counterexample would be the anti-derivative of x+5, which is (1/2)x^2 + 5x , just like in the example they use x^2 (which is the anti-derivative of 2x) From there, this is what I did:
if f(x) = (1/2)x^2 + 5x
then --> d/dx f(x+5) = d/dx [ (x^2 + 20x +75)/2 ] = (4x+40)/4
but on the other hand f'(x) = x +5
so f'(x+5) = x + 5
therefore x + 5 ≠ (is not equal to) (4x+40)/4
--------------------------------------------------------------------
Similarly, I applied the same idea for question (b)
if g(t) = (1/6)t^2
then --> d/dt g(t/3) = d/dt [ (t^2)/54 ] = (108t)/54^2
but on the other hand f'(x) = t/3
so f'(t/3) = (36t)/18^2
therefore (36t)/18^2 ≠ (is not equal to) (108t)/54^2
I think these are not the counterexamples they are asking for. How is the right way to find them? Should I use other equations or formulas?
Thank you in advance for any help you might be able to provide.
Sincerely
megadeth95