Derivative of k(x): Understand with Examples

  • Thread starter Thread starter Tom McCurdy
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the derivatives of the function y=k(x) under specific transformations. When the derivative of y=k(x) equals 2 at x=1, the derivatives for the transformations k(2x), k(x+1), and k(x/4) are determined using the chain rule. The results are 4 for k(2x) at x=1/2, 2 for k(x+1) at x=0, and 1/2 for k(x/4) at x=4. The correct application of the chain rule is emphasized as crucial for these calculations.

PREREQUISITES
  • Understanding of derivatives and their notation
  • Familiarity with the chain rule in calculus
  • Basic knowledge of function transformations
  • Ability to evaluate functions at specific points
NEXT STEPS
  • Study the chain rule in detail, focusing on its application in various scenarios
  • Practice derivative calculations for composite functions
  • Explore function transformations and their effects on derivatives
  • Learn about higher-order derivatives and their significance
USEFUL FOR

Students learning calculus, educators teaching derivative concepts, and anyone seeking to strengthen their understanding of function transformations and the chain rule.

Tom McCurdy
Messages
1,021
Reaction score
1
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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K