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# Differentiation question watch

1. Hi,

Can some please help me with this question. I've been working on it for ages but just don't understand!

I need to differentiate the following equation:

Y = L + M e^(-xt)

L = 40
M = 10
X = 3
t = 4
Is it something like dy/dx = 0 + -XMe^(-xt) or am I completely wrong?
2. (Original post by YYY)
Hi,

Can some please help me with this question. I've been working on it for ages but just don't understand!

I need to differentiate the following equation:

Y = L + M e^(-xt)

L = 40
M = 10
X = 3
t = 4
Is it something like dy/dx = 0 + -XMe^(-xt) or am I completely wrong?
Are there any variables in this equation? If so, which are they?

If it's x, then it'd be dy/dx = -tMe-xt.

If it's t, then it'd be dy/dt = -xMe-xt.

I assume you'd then put your values in to find the value of dy/dx or dy/dt at this point.
3. (Original post by Gaiaphage)
Are there any variables in this equation? If so, which are they?

If it's x, then it'd be dy/dx = -tMe-xt.

If it's t, then it'd be dy/dt = -xMe-xt.

I assume you'd then put your values in to find the value of dy/dx or dy/dt at this point.
Thanks for your reply. The variables are t. So I'm right but just need to put in dy/dt, not dy/dx?

Is the L equal to 0 when differentiated?
4. (Original post by YYY)
Thanks for your reply. The variables are t. So I'm right but just need to put in dy/dt, not dy/dx?

Is the L equal to 0 when differentiated?
Yeah, if t is a variable and x isn't then you'd do dy/dt. That gives dy/dt = -xMe-xt and you can put your values in from there.

As L is a constant then yes it goes to 0 (you can think of that by thinking of the gradient of a line y = 3, the gradient would be 0)
5. (Original post by Gaiaphage)
Yeah, if t is a variable and x isn't then you'd do dy/dt. That gives dy/dt = -xMe-xt and you can put your values in from there.

As L is a constant then yes it goes to 0 (you can think of that by thinking of the gradient of a line y = 3, the gradient would be 0)

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