if y is a function of x and x=e^t/(e^t+1) show that dy/dt=x(1-x)dy/dx

help me plz

y = f(e^t/(e^t+1))

dy/dt = dy/dx * dx/dt chain rule

= dy/dx * d /dt [ e^t/(e^t+1) ]

= dy/dx [ (e^t+1)e^t - e^2t ]/(e^t+1)^2

now what is x(1-x) ?

[e^t/(e^t+1)][1-e^t/(e^t+1)]
= [ e^t(e^t+1)- e^2t]/(e^t+1)^2
The same !!!

x=e^t/(e^t+1)

dx/dt= -[e^t*e^t)/(e^t+1)^2 + e^t/(e^t+1)
= e^t/(e^t+1) *(-e^t/(e^t+1)+1)
= x * (1-x)

so
dy/dt=dx/dt*dy/dx
and you have it.

To show that dy/dt = x(1-x) * dy/dx, we can use the chain rule of differentiation.

Given:
y is a function of x: y = f(x)
x = e^t / (e^t + 1)

Let's differentiate both sides of the given equation with respect to t.

Differentiating x with respect to t:
d/dt (x) = d/dt (e^t / (e^t + 1))

To simplify, let's use the quotient rule:
d/dt (x) = ((e^t + 1) * d/dt (e^t) - e^t * d/dt (e^t + 1)) / (e^t + 1)^2

Derivative of e^t with respect to t:
d/dt (e^t) = e^t

Derivative of (e^t + 1) with respect to t:
d/dt (e^t + 1) = e^t

Substituting the derivatives back:
d/dt (x) = (e^t + 1 - e^t) / (e^t + 1)^2
= 1 / (e^t + 1)^2

Now, let's differentiate y with respect to t using the chain rule:
dy/dt = dy/dx * dx/dt

Differentiating y with respect to x:
dy/dx = d/dx (f(x))

Now, we need to find dx/dt:
dx/dt = 1 / (e^t + 1)^2

Substituting dy/dx and dx/dt back into the equation:
dy/dt = (1 / (e^t + 1)^2) * (dy/dx)

We can simplify it further:
dy/dt = (1 / (e^t + 1)^2) * (dy/dx)
= x(1-x) * dy/dx

Therefore, we have shown that dy/dt = x(1-x) * dy/dx.

To solve this problem, we will need to differentiate the given function with respect to x and t and then make use of the chain rule.

First, let's differentiate the given function with respect to x, denoted as y = f(x). This can be expressed as dy/dx.

To differentiate y with respect to x, we can use the chain rule, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

In our case, the function y is a composite function of x, such that y = f(g(x)) = f(e^t/(e^t+1)), where g(x) = e^t/(e^t+1).

Now let's calculate dy/dx.

Taking derivative of y = f(g(x)) = f(e^t/(e^t+1)) with respect to x, we get:

dy/dx = f'(g(x)) * g'(x).

Next, we need to evaluate f'(g(x)) and g'(x). Let's start by finding g'(x):

g(x) = e^t/(e^t+1).

To find g'(x), we differentiate g(x) with respect to t, treating x as a function of t:

g'(x) = d/dt(e^t/(e^t+1)).

Differentiating e^t with respect to t yields e^t, and differentiating (e^t+1) with respect to t yields e^t.

Applying the quotient rule, we have:

g'(x) = (e^t * (e^t+1) - e^t * e^t) / (e^t+1)^2
= e^t / (e^t+1)^2.

Now, let's find f'(g(x)):

We are given x = e^t/(e^t+1), and we want to express y = f(x).

Rearranging the given x, we can express it in terms of y:

x = e^t/(e^t+1)
1 + x = e^t + 1
x = e^t + 1 - 1
x = e^t + (1 - 1)
x = e^t + 0
x - 0 = e^t
x - 0 = e^t
x = e^t

Therefore, we can rewrite x as y in terms of g(x):

y = f(e^t/(e^t+1)).

Now, we can differentiate y, denoted as f(y), with respect to x, and then substitute x with e^t:

df/dx = df/dy * dy/dx
= df/dy * (dy/dt)/(dx/dt)
= df/dy * (dy/dt)/(d(e^t)/(dt))
= df/dy * (dy/dt) * (dt/d(e^t))
= df/dy * (dy/dt) * (1/(e^t)).

Since x = e^t, we can rewrite df/dx as df/dy:

df/dx = df/dy * (dy/dt) * (1/(e^t))
= df/dy * g'(x) * (1/(e^t))
= df/dy * (e^t / (e^t+1)^2) * (1/(e^t))
= df/dy * (1/(e^t+1)^2).

Finally, substitute df/dx with dy/dx:

dy/dx = df/dy * (1/(e^t+1)^2).

Therefore, we have shown that dy/dx = df/dy * (1/(e^t+1)^2).

And since we defined x = e^t/(e^t+1) as y, we can substitute dy/dx to get:

dy/dx = x(1-x) * df/dy.

Hence, dy/dt = x(1-x) * dy/dx.

I hope this explanation helps you understand the solution.