- #1

- 125

- 0

## Main Question or Discussion Point

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:

2x

if we go (d/dx)(2x

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)(2x

Now (d/dt)(2x

Right?

Ok so now what if we have

y=2x

x=4t

(dy/dx)=4x

(d/dx)(2x

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)

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?

~~~~~~~~~~~~~~~~

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?

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:

2x

^{2}+3xif we go (d/dx)(2x

^{2}+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)(2x

^{2}+3x) still (4x+3)? after all, we are only asking about the differential in terms of xNow (d/dt)(2x

^{2}+3x) = (d/dx)(dx/dt)(2x^{2}+3x) = (4x+3)(2)Right?

Ok so now what if we have

y=2x

^{2}x=4t

(dy/dx)=4x

(d/dx)(2x

^{2})=4xAlso, (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}=32t^{2}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?

~~~~~~~~~~~~~~~~

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?

Last edited: