Derivative of k(x): Understand with Examples

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The discussion centers on finding the derivatives of k(2x), k(x+1), and k(x/4) given that the derivative of y=k(x) equals 2 when x=1. The first participant initially misapplies the chain rule, leading to incorrect calculations for the derivatives. Clarifications reveal that for k(2x), the correct derivative is 4, while for k(x+1) it is 2, and for k(x/4) it is 1/2. The conversation emphasizes the importance of correctly applying the chain rule and substitution in derivative calculations. Understanding these principles is crucial for solving similar problems effectively.
Tom McCurdy
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If the derivative of y=k(x) equals 2 when x=1 what is the derivative of

a. k(2x) when x = 1/2
b. k(x+1) when x=0
c. k(x/4) when x =4

I had two different ideas both seemed wrong
the first idea was to implore some type of chain rule so
a. 2*2=4
b. 2*1=2
c 2*1/4= 1/2

by saying the outside function would be 2 and then taking the derivative of the inside function, but you can't really do this so it seems wrong

the second idea seems even more wrong to simpliy before the derivative so that everything would come to be k(x) making a,b, and c =2

if someone could help me that would be great

thx
tom
 
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this problem seems as straightforward as your second idea i think..

i mean if you evaluate something at f(1) or f(2/2) its the same thing..

but then again, why would they ever ask that if this is the case?


Josh
 
Tom McCurdy said:
If the derivative of y=k(x) equals 2 when x=1 what is the derivative of
a. k(2x) when x = 1/2
b. k(x+1) when x=0
c. k(x/4) when x =4
I had two different ideas both seemed wrong
the first idea was to implore some type of chain rule so
"employ", not "implore" (unless you are begging the chain rule to work!)
a. 2*2=4
b. 2*1=2
c 2*1/4= 1/2
by saying the outside function would be 2 and then taking the derivative of the inside function, but you can't really do this so it seems wrong
The "outside function" is not 2- the inside function is 2x.
The derivative of k(2x) is k'(2x)(2) where "k'(2x)" means just the derivative of k(x) evaluated at 2x. Since y'(1)= 2, y'(2(1/2))= y'(1)= 2 and so
the derivative of k(2x) at x= 1/2 is 2(2)= 4.

the second idea seems even more wrong to simpliy before the derivative so that everything would come to be k(x) making a,b, and c =2
if someone could help me that would be great
thx
tom
If by "simplify" you mean "substitute", yes, that's a great idea. You should eventually learn to apply the chain rule quickly without having to write down the substitution but it's good practice to write it out while you are still learning.
In (a), let u= 2x. Then dk/dx= (dk/du)(du/dx). du/dx= 2 and u= 1 when x= 1/2 so the derivative is (2)(2)= 4.

In (b), let u= x+ 1. Now du/dx= 1 and u= 1 when x= 0 so the derivative is
(2)(1)= 2.

In (c), let u= x/4. Now du/dx= 1/4 and u= 1 when x= 4 so the derivative is
(2)(1/4)= 1/2.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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