Why Is the Chain Rule Not Used in Differentiating h(x) = 3f(x) + 8g(x)?

member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For part(a),
1683504334746.png

The solution is,
1683504351004.png

However, why do they not take the derivative of the inner function (if it exists) of f(x) or g(x) using the chain rule? For example if ##f(x) = \sin(x^2)##

Many thanks!
 
Physics news on Phys.org
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For part(a),
View attachment 326130
The solution is,
View attachment 326131
However, why do they not take the derivative of the inner function (if it exists) of f(x) or g(x) using the chain rule? For example if ##f(x) = \sin(x^2)##

Many thanks!
There is no inner function. The chain rule is for a composition of functions, like f(g(x)). That does not appear in this problem. The derivative is with respect to x and both f(x) and g(x) are direct functions of x.
 
  • Like
Likes member 731016
ChiralSuperfields said:
However, why do they not take the derivative of the inner function (if it exists) of f(x) or g(x) using the chain rule?
As already noted, there is no "inner function," but the derivative of h(x) (i.e., h'(x)) requires only the use of the sum rule and constant multiple rules for derivatives. Thus h'(x) = 3f'(x) + 8g'(x). From the given information it's easy to calculate h'(4).

BTW, you don't take "h'(x) of h(x)" similar to what you have in the title. You can find the derivative of h(x) or differentiate h(x).
 
  • Like
Likes member 731016
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top