Leibniz notation when taking derivatives

[dumbQuestion]

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?

 

[HallsofIvy]

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.

 

[dumbQuestion]

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.

 

[dumbQuestion]

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)

?

 

[Bipolarity]

d/du(F(u))=d/du((3t+4)²)=d/du(u²)=2u=2(3t+4)=f(t)=f(u)

?

The last statement $= f(t) = f(u)$ 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