Understanding the Meaning of Derivatives and Substituting with Variable p

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SUMMARY

The discussion clarifies the relationship between derivatives when substituting a variable p for the derivative of a function y(x). Specifically, it establishes that the expression dp/dy is a shorthand for the chain rule, indicating differentiation with respect to y. The correct interpretation is that p is a function of y, represented as p(y(x)) = y'(x). The final relationship derived is dp/dy = y''(x)/y'(x), demonstrating how the second derivative relates to the first derivative in this context.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives.
  • Familiarity with the chain rule in differentiation.
  • Knowledge of function notation and variable substitution.
  • Ability to interpret and manipulate mathematical expressions involving derivatives.
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  • Study the chain rule in calculus to deepen understanding of variable substitution.
  • Explore the relationship between first and second derivatives in detail.
  • Practice problems involving function substitution and differentiation.
  • Learn about implicit differentiation and its applications in calculus.
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why p''=pp' ??

i have a function
y(x)

i want to substitute by variable p.

p=y'(x)=dy/dx

every derivative is by dx because its the smallest variable
so why

why dp/dy=p'


i can't see the meaning of derivative by y

??
 
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First of all, writing dp/dy=p' implies that you're differentiating with respect to y. Otherwise, you are right, dp/dy is not equal to dp/dx in this case.

The expression [tex]\frac{dp}{dy}[/tex] is really just a short-hand for the chain rule. The best thing to do here is not to think of p as a variable but instead as a function p(y(x))=y'(x). What it's asking is "What happens to the value of p(y(x)) as the value of y(x) changes?" That is, we want to find
[tex]\frac{{\rm change\ in\ }p\bigl(y(x)\bigr)}{{\rm change\ in\ }y(x)},[/tex]​
or, in "semi-short-hand" notation,
[tex]\frac{dp\bigl(y(x)\bigr)}{dy(x)},[/tex]​
and in standard short-hand notation,
[tex]\frac{dp}{dy}.[/tex]​
Now, since both p(y(x)) and y(x) are in terms of x, we can re-write these (albeit with a slight abuse of notation) as
[tex] \begin{align*}<br /> \frac{dp}{dy} &=\frac{dp\bigl(y(x)\bigr)}{dy(x)} \\<br /> &= \frac{dp\bigl(y(x)\bigr)}{dy(x)} \cdot \frac{dx}{dx} \\<br /> &= \frac{dp\bigl(y(x)\bigr)}{dx} \biggl/ \frac{dy(x)}{dx}\\<br /> &= \frac{dp}{dx}\biggl/\frac{dy}{dx}\\<br /> &= \frac{y''(x)}{y'(x)}.<br /> \end{align*}[/tex]​
 

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