Leibniz notation when taking derivatives

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  • #1
dumbQuestion
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Hello,I am encountering some major confusion. When taking just garden variety f(x)=y derivatives of the form dy/dx, I don't encounter any problems. But recently I started taking derivatives of parametric equations, or switching things up using polar equations and I realized perhaps I'm not so solid on these things as I once thought.Is there any way someone could tell me if I'm right about a few problems?So say we have:

2x2+3x

if we go (d/dx)(2x2+3x) the answer is just (4x+3)Now what if x is actually a function of another variable, say, t for instance. Say we know in reality, that x=2t.Now is (d/dx)(2x2+3x) still (4x+3)? after all, we are only asking about the differential in terms of xNow (d/dt)(2x2+3x) = (d/dx)(dx/dt)(2x2+3x) = (4x+3)(2)

Right?Ok so now what if we havey=2x2
x=4t

(dy/dx)=4x

(d/dx)(2x2)=4x

Also, (dx/dt)=4, which is the same as writing (d/dt)(4t)=4Now, we could also write:

(dy/dx)=(dy/dt)/(dt/dx)If we plug in (4t) for x, we get
y=2(4t)2=32t2

So dy/dt=64t

We know (dx/dt)=4 --> (dt/dx)=1/4 (since this denominator never goes to 0)

so (dy/dx) = (dy/dt)*(dt/dx)=64t*(1/4) = 16tFrom the equation x=4t given above we know, t = (1/4)x

so (dy/dx)= 16t= 16(x/4) = 4xIs this all right?~~~~~~~~~~~~~~~~Another confusing question for me:say we have x=3t

Would d/dx(sinx)=cosx

Or would it be

d/dx(sinx)=(cosx)(x')=(cosx)(dx/dt)=(cosx)(3)When I see d/dx, I always think that just means, "take the derivative of this in terms of x only". What if we know x is a function of another variable though?
 
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  • #2
dumbQuestion said:
Hello,


I am encountering some major confusion. When taking just garden variety f(x)=y derivatives of the form dy/dx, I don't encounter any problems. But recently I started taking derivatives of parametric equations, or switching things up using polar equations and I realized perhaps I'm not so solid on these things as I once thought.


Is there any way someone could tell me if I'm right about a few problems?


So say we have:

2x2+3x

if we go (d/dx)(2x2+3x) the answer is just (4x+3)
Yes, that is correct.

Now what if x is actually a function of another variable, say, t for instance. Say we know in reality, that x=2t.


Now is (d/dx)(2x2+3x) still (4x+3)? after all, we are only asking about the differential in terms of x
Yes, that is still correct. We are differentiating with respect to x so any dependence on t is irrelevant.

Now (d/dt)(2x2+3x) = (d/dx)(dx/dt)(2x2+3x) = (4x+3)(2)

Right?
The middle term there looks suspicious. You should have (dx/dt)(d/dx)(2x2+3x)
so it doesn't look like you want to differentiate (dx/dt) with respect to x. Other than that, yes, that is a correct application of the "chain rule". Notice that could also have replaced x with 2t:
2x2+ 3x= 2(2t)2+ 3(2t)= 8t2+ 6t and the derivative of that is 16t+ t. Of course, that is the same as 2(4x+ 3)= 2(4(2t)+ 3)= 16t+ 6

Ok so now what if we have


y=2x2
x=4t

(dy/dx)=4x

(d/dx)(2x2)=4x

Also, (dx/dt)=4, which is the same as writing (d/dt)(4t)=4


Now, we could also write:

(dy/dx)=(dy/dt)/(dt/dx)


If we plug in (4t) for x, we get
y=2(4t)2=32t2

So dy/dt=64t

We know (dx/dt)=4 --> (dt/dx)=1/4 (since this denominator never goes to 0)

so (dy/dx) = (dy/dt)*(dt/dx)=64t*(1/4) = 16t


From the equation x=4t given above we know, t = (1/4)x

so (dy/dx)= 16t= 16(x/4) = 4x


Is this all right?
Yes, that is completely correct.
 
  • #3
Thanks again, hallsOfIvy. Since I have come to this forum you have answered so so many of my questions. I almost feel bad asking questions. I just hope you know how much I appreciate yours and others contributions. For myself I am just reviewing all of these concepts which are old to me, and it has helped my review so incredibly much to get feedback and insight from users here. I wish I could contribute more so I wouldn't just be taking taking taking and not giving, but unfortunately I am not yet confident enough in any answers to think I would be giving a correct answers to users who are submitting questions.
 
  • #4
Perhaps if I could ask one more question to make sure I've "got it"So, say we have a function F(t). We can say F(t) is an antiderivative of f(t) if d/dt(F(t))=f(t)Now if we have two such functions, does that mean for some u we can automatically say that:

F(u) is an antiderivative of u, and that d/du(F(u))=f(u) ?For example,

for f(x)=2x, we know that F(x)=x2 is an antiderivative, and that d/dx(F(x))=d/dx(x2)=2x=f(x)Now let's say, u=3t+4

Does this mean that if we plug u in, and get

f(u)=2(3t+4) and F(u)=(3t+4)2, that F(u) is still an antiderivative of f(u) and therefored/du(F(u))=d/du((3t+4)2)=d/du(u2)=2u=2(3t+4)=f(t)=f(u)

?
 
  • #5
dumbQuestion said:
d/du(F(u))=d/du((3t+4)2)=d/du(u2)=2u=2(3t+4)=f(t)=f(u)

?

The last statement [itex] = f(t) = f(u) [/itex] is incorrect, f(t) does not equal f(u).
f(u) equals f(3t+4) which is not the same thing as f(t).

Everything else looks good though.

BiP
 

1. What is Leibniz notation when taking derivatives?

Leibniz notation, also known as the "prime notation", is a mathematical notation used to represent the derivative of a function. It is written as f'(x) or dy/dx, where f(x) is the function and x is the independent variable.

2. Why is Leibniz notation used for derivatives?

Leibniz notation is used because it is a clear and concise way to represent the derivative of a function. It also allows for easy manipulation and calculation of derivatives using the rules of differentiation.

3. How do you read Leibniz notation?

The notation f'(x) or dy/dx can be read as "the derivative of f with respect to x" or "the derivative of y with respect to x". This indicates that the function is being differentiated with respect to the specified variable.

4. What is the difference between Leibniz notation and Newton's notation for derivatives?

The main difference between Leibniz notation and Newton's notation is the way derivatives are represented. Newton's notation uses the symbol d followed by the function and independent variable, while Leibniz notation uses the prime symbol ' or the fraction notation dy/dx. Both notations are equivalent and can be used interchangeably.

5. Can Leibniz notation be used for higher order derivatives?

Yes, Leibniz notation can be used for higher order derivatives, which represent the rate of change of a function's rate of change. The notation for second order derivative is written as f''(x) or d2y/dx2, and for third order derivative it is written as f'''(x) or d3y/dx3, and so on.

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