1Mperios
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Hello,

If dy/dx means decreasing the number of exponent for example x^2 = dy/dx 2x

Is 1^1 then not 1 x 1^0, and 1^0 = 1

dy/dx 1 x 1 = 1???

can someone please explain what differentiation means?

thanks
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SimonM
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#2
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\frac{d}{dx} c = 0 for any constant c
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okay_kay
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its just 0
if y=1... same as y=1 x x^0... then dy/dx= 0 x x^-1... and anything multiplied by 0 is 0
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Siddd
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dy/dx of 1 is 0
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P34c0ck1991
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your finding the gradient of something, and you can use a point to find the gradient of each point of a graph, the gradient if x^2 is 2x, so at (1,1) its a gradient of 2
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Octavian
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No doubt you will get a more full explanation, but the derivative of a function is the gradient of the line, as you should know. If your line has equation x=1, or any number, the gradient is just 0. Try not to think of the rules for polynomials as a definition, more as handy ways to differentiate polynomials.
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~Viola~
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Isn't it just nothing?

It finds the adjacent line on a round shaped graph.

If you had

3x(power of 3) + 2x (power of 2) + 6

dy/dx = 9x (power of 2) + 4x

take the power, times it with the number before the x then take 1 away from the power. Numbers without an x disappear all together
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corney91
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0. Think of it as the gradient of y=1
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gyyy2807
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#9
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it's just zero. when differentiating any number, you just lose that number.
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Xei
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#10
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1 = 1*x^0.

When you differentiate any variable you multiply by the power of the variable and decrease the power by 1, so:

d1/dx = 0*1*x^-1 = 0

For any constant a you can write it as ax^0, and differentiating that with regards to x with involve multiplying by 0.
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1Mperios
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(Original post by Xei)
1 = 1*x^0.

When you differentiate any variable you multiply by the power of the variable and decrease the power by 1, so:

d1/dx = 0*1*x^-1 = 0

For any constant a you can write it as ax^0, and differentiating that with regards to x with involve multiplying by 0.
why is 1 = 1x^0?
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Zhen Lin
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x^0 = 1 for any x \ne 0. (And if we want to differentiate it, we will usually assume x^0 = 1 for all x.) So 1 x^0 = 1.
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BJP
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It's just 0. When differentiating a constant it just disappears. E.g. differentiating 2x+1 gives 2.
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Xei
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why is 1 = 1x^0?
Any variable to power 0 is 1. This is because:

a^p * a^q = a^p+q

a^0 * a^q = a^q+0 = a^q

As multiplying by a^0 has no effect, a^0 must be 1.

You would write 1 as 1*x^0 because you're differentiating with respects to x.

The thing you wrote, about writing 1^1 and then using the differentiating rule on it doesn't work, because we're not differentiating with respects to 1, we're differentiating with respects to x. You only use that rule on powers of x.
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122025278
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(Original post by 1Mperios)
Hello,

If dy/dx means decreasing the number of exponent for example x^2 = dy/dx 2x

Is 1^1 then not 1 x 1^0, and 1^0 = 1

dy/dx 1 x 1 = 1???

can someone please explain what differentiation means?

thanks
Calculus is the association of change in one variable with respect to another

So dy/dx literally means how the variable y changes as x changes.

Imagine a graph, draw the line y = 1. It doesn't matter what value of x you look at, y = 1. It x changes, decreases or increaes, y will always be 1 won't it.

So dy/dx = 0 because no matter how x changes, y doesn't change. So there is ZERO change in y as x changes, because y always remains at 1 doesn't it.
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1Mperios
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(Original post by 122025278)
Calculus is the association of change in one variable with respect to another

So dy/dx literally means how the variable y changes as x changes.

Imagine a graph, draw the line y = 1. It doesn't matter what value of x you look at, y = 1. It x changes, decreases or increaes, y will always be 1 won't it.

So dy/dx = 0 because no matter how x changes, y doesn't change. So there is ZERO change in y as x changes, because y always remains at 1 doesn't it.
right, this is one line, what happens if u have y = 3 at one point and y = -3 at another point, it is still a line and x changes....
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Oh I Really Don't Care
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(Original post by 1Mperios)
right, this is one line, what happens if u have y = 3 at one point and y = -3 at another point, it is still a line and x changes....
... huh
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benwellsday
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(Original post by 1Mperios)
right, this is one line, what happens if u have y = 3 at one point and y = -3 at another point, it is still a line and x changes....
The rest of the posts explain what you've done wrong but what I've noticed is you've said dy/dx reduces the exponent. Are you doing C1 by any chance, as this is the main rule they teach you in there about differentiation, but you've got it the wrong way.

dy/dx does reduce the exponent, but only of x, like if y= x^3, dy/dx = 3x^2, but if y = 2^3, 2 isn't x, so you don't reduce any exponents or anything. If you have some constant, as said, like y=2^3, then dy/dx = 0, and if you have something like  y = 2^3x^3 then dy/dx =  3(2^3)x^2 the 2^3 just acts like a multiple. This might become more clear from the next bit...

Differentiation, at this level, means the gradient of a graph, so if you had y = x^3 + 4x - 3, you could find the gradient at any point. This is useful, which is why differentiation is important, for example you can find the equation of a tangent line to any point on the graph, or you can find (local) maximums and minimums.
Now back to your example, y = 1, imagine this as a graph, it's just a straight horizontal line. To find the gradient you draw a triangle, but in this case you can't because the graph is flat, but you can think of it kind of like this, y stays the same, and x changes alot, hence the change in y is 0, and the change in x is anything (but not 0, that'd just be annoying for the next bit), so the gradient is 0 / (change in x) = 0. This is why if y=1, dy/dx = 0.

Now for the last bit you said, y = 3 at one point and y = -3 at another point, this is a line, but you have picked two points, what you need is the equation of this line to fdo differentiation. This could be anything though, so think of it as the case y=x. When x = 3, y=3, and when x=-3, y=-3. But when you differentiate, i.e. do dy/dx, you have to look at the y = x part. After you've differentiated you can substitute 3 or -3 for every x in the equation. This example is boring, you get 1 no matter what. So here's another

y= x^2
Find dy/dx at 3.
First, differentiate x^2, dy/dx = 2x.
Now substitute in x = 3, so dy/dx = 6.
If you'd done this the other way round it'd go wrong, i.e. if you said x = 3, so y = 3, so dy/dx = 0, then that is wrong, you can't do it in that order.

Hopefully in all those points at least one of them is the problem you're having.
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1Mperios
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#19
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(Original post by benwellsday)
The rest of the posts explain what you've done wrong but what I've noticed is you've said dy/dx reduces the exponent. Are you doing C1 by any chance, as this is the main rule they teach you in there about differentiation, but you've got it the wrong way.

dy/dx does reduce the exponent, but only of x, like if y= x^3, dy/dx = 3x^2, but if y = 2^3, 2 isn't x, so you don't reduce any exponents or anything. If you have some constant, as said, like y=2^3, then dy/dx = 0, and if you have something like  y = 2^3x^3 then dy/dx =  3(2^3)x^2 the 2^3 just acts like a multiple. This might become more clear from the next bit...

Differentiation, at this level, means the gradient of a graph, so if you had y = x^3 + 4x - 3, you could find the gradient at any point. This is useful, which is why differentiation is important, for example you can find the equation of a tangent line to any point on the graph, or you can find (local) maximums and minimums.
Now back to your example, y = 1, imagine this as a graph, it's just a straight horizontal line. To find the gradient you draw a triangle, but in this case you can't because the graph is flat, but you can think of it kind of like this, y stays the same, and x changes alot, hence the change in y is 0, and the change in x is anything (but not 0, that'd just be annoying for the next bit), so the gradient is 0 / (change in x) = 0. This is why if y=1, dy/dx = 0.

Now for the last bit you said, y = 3 at one point and y = -3 at another point, this is a line, but you have picked two points, what you need is the equation of this line to fdo differentiation. This could be anything though, so think of it as the case y=x. When x = 3, y=3, and when x=-3, y=-3. But when you differentiate, i.e. do dy/dx, you have to look at the y = x part. After you've differentiated you can substitute 3 or -3 for every x in the equation. This example is boring, you get 1 no matter what. So here's another

y= x^2
Find dy/dx at 3.
First, differentiate x^2, dy/dx = 2x.
Now substitute in x = 3, so dy/dx = 6.
If you'd done this the other way round it'd go wrong, i.e. if you said x = 3, so y = 3, so dy/dx = 0, then that is wrong, you can't do it in that order.

Hopefully in all those points at least one of them is the problem you're having.
loool nah I am currently doing C4,I knew what the answer is but I was just curious about this... Thank you for your good explanation.
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nuodai
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\displaystyle \frac{\mbox{d}y}{\mbox{d}x} is just used to represent the gradient of a graph. If you draw the graph of y = 1, you'll notice that it's a straight horizontal line, so its gradient (and hence the value of \displaystyle \frac{\mbox{d}y}{\mbox{d}x} is zero. The same is true for any constant.

Another way is interpreting it as a polynomial (where the constant is a term in x^0); that is, 1 = a_0x^0 + a_1x^1 + a_2x^2 + \cdots + a_nx^n. Because on the left hand side there are no terms in x, x^2, \dots, x^n, you can ignore those, giving you:

\newline\displaystyle a_0x^0 = 1 \Rightarrow \frac{d}{dx}\left( a_0x^0 \right) = 0\cdot a_0\cdot x^{-1} = 0.
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