Are Variables and Constants Distinguishable in Differentiation?

opticaltempest
Messages
135
Reaction score
0
Before we differentiate, we must know whether a variable in that expression represents a constant or variable, correct?

For example, if we have the function,

f(x) = r^3 x^2

f'(x) = \frac{d}{{dx}}\left[ {r^3 x^2 } \right]​

(1) Now if r represents a variable then,

f&#039;(x) = r^3 \cdot \frac{d}{{dx}}\left[ {x^2 } \right] + \frac{d}{{dx}}\left[ {r^3 } \right] \cdot x^2<br />

f&#039;(x) = 2r^3 x + 3r^2 x^2​

But if we know r to be a constant then,

f&#039;(x) = 2r^3 x​

So it seems like it is very important to know exactly what the variable represents in a function. Am I correct? In my current calculus textbook, I don't think this topic was covered thoroughly enough.

EDIT: On line labeled (1) the word "constant" was replaced by the intented word, "variable". My mistake...
 
Last edited:
Physics news on Phys.org
Now if r represents a function of x then,

f&#039;(x) = r^3 \cdot \frac{d}{{dx}}\left[ {x^2 } \right] + \frac{d}{{dx}}\left[ {r^3 } \right] \cdot x^2<br />

which, by the chain rule, is

f&#039;(x) = 2r^3 x + 3r^2 r^{\prime} x^2​

But if we know r to be a constant then, r^{\prime}=0

f&#039;(x) = 2r^3 x​
 
Could you elaborate more on what is meant by "r is a function of x"?

Is there a difference in me saying f(x) vs f(r,x) ?

Since I said f(x) can I think of it like,

"r doesn't allow for any inputs, only x allows inputs. Therefore the value of r will depend upon
what value of x you put in." ?

By me saying f(x)=r*x does it mean that r MUST be some defined constant since it cannot have an input?
 
Last edited:
If r represents a variable, indpendent of x, then it would make no sense to write f(x)= r3x2- it would have to be f(x,r)= r3x2.

If you suspect that r is a function of x (for example that r= cos(x) so that f(x)= r3x2= cos3(x)x2 and so f'(x)= -3x2sin2(x)+ 2xcos3(x)), then you can "cover all bets" by writing f'(x)= 3r2x2r'(x)+ 2r3x. In that case, if it happens that r is a constant, r'(x)= 0 giving you f'(x)= 2r3x which is correct.

Generally, whether or not a letter represents a variable or a constant should be clear from the context.
 
Last edited by a moderator:
Translation: if something isn't what we supposed it was then its behaviour may change? Answer: Yes. That has nothing to do with maths though.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top