x = e^t

y = te^-t

dy/dx = y'/x'

= (1-t)e^-t/e^t

= (1-t)e^-2t

d^2/dx^2 = (x'y"-x"y')/x'^3

= ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/e^3t

= (2t-3)e^-3t

or, do it directly

x = e^t, so

y = lnx/x

y' = (1-lnx)/x^2 = (1-t)e^-2t

y" = (2lnx-3)/x^3 = (2t-3)e^-3t

# Find

dy/dx and d2y/dx2.

x = et, y = te−t

Steve
answered

10 years ago

10 years ago

sara
answered

10 years ago

10 years ago

thanks. so how woudl i find For which values of t is the curve concave upward?

Steve
answered

10 years ago

10 years ago

just as usual. Where is y" positive?

e^-3t is always positive, so all you need is

2t-3 > 0

t > 3/2

e^3/2 = 4.48, and you can see from the graph that it is concave upward for x > 4.48

http://www.wolframalpha.com/input/?i=plot+x+%3D+e^t%2C+y+%3D+te^-t+for+1%3C%3Dt%3C%3D2

e^-3t is always positive, so all you need is

2t-3 > 0

t > 3/2

e^3/2 = 4.48, and you can see from the graph that it is concave upward for x > 4.48

http://www.wolframalpha.com/input/?i=plot+x+%3D+e^t%2C+y+%3D+te^-t+for+1%3C%3Dt%3C%3D2

Clown Bot
answered

1 year ago

1 year ago

Why did the mathematician take up juggling? Because he wanted to find the derivative of x = et and y = te-t!

Now, let's actually find the derivatives. We'll start with the first derivative, dy/dx.

To find dy/dx, we use the chain rule. The chain rule states that if we have a function y(t) and another function x(t), then dy/dx = (dy/dt) / (dx/dt).

So, let's find dy/dt first. The derivative of y = te-t with respect to t can be found using the product rule. Let's call u = t and v = e-t.

Using the product rule, du/dt = 1 and dv/dt = -e-t.

Now, we can apply the product rule: dy/dt = u * dv/dt + v * du/dt = t * (-e-t) + e-t * 1.

Simplifying this, we get dy/dt = -te-t + e-t.

For dx/dt, notice that x = et. The derivative of x with respect to t is simply dx/dt = e-t.

Finally, we can find dy/dx by dividing dy/dt by dx/dt:

dy/dx = (dy/dt) / (dx/dt) = (-te-t + e-t) / e-t.

Simplifying this, we get dy/dx = -t + 1.

Now, onto the second derivative, d2y/dx2. But let's save that for another joke! Why did the mathematician become a clown? Because he got tired of all these derivatives and needed a change!

I hope this mathematical comedy routine helps you find the derivatives you were looking for!

Now, let's actually find the derivatives. We'll start with the first derivative, dy/dx.

To find dy/dx, we use the chain rule. The chain rule states that if we have a function y(t) and another function x(t), then dy/dx = (dy/dt) / (dx/dt).

So, let's find dy/dt first. The derivative of y = te-t with respect to t can be found using the product rule. Let's call u = t and v = e-t.

Using the product rule, du/dt = 1 and dv/dt = -e-t.

Now, we can apply the product rule: dy/dt = u * dv/dt + v * du/dt = t * (-e-t) + e-t * 1.

Simplifying this, we get dy/dt = -te-t + e-t.

For dx/dt, notice that x = et. The derivative of x with respect to t is simply dx/dt = e-t.

Finally, we can find dy/dx by dividing dy/dt by dx/dt:

dy/dx = (dy/dt) / (dx/dt) = (-te-t + e-t) / e-t.

Simplifying this, we get dy/dx = -t + 1.

Now, onto the second derivative, d2y/dx2. But let's save that for another joke! Why did the mathematician become a clown? Because he got tired of all these derivatives and needed a change!

I hope this mathematical comedy routine helps you find the derivatives you were looking for!

Explain Bot
answered

1 year ago

1 year ago

To find dy/dx and d2y/dx2, we need to use the chain rule and the product rule.

Given that:

x = et

y = te−t

First, let's compute dy/dx:

To find dy/dx, we will differentiate y with respect to t and then multiply it by dt/dx.

Differentiating y with respect to t:

dy/dt = d(te−t)/dt

Using the product rule, we have:

dy/dt = t(d(e−t)/dt) + (te−t)(d(t)/dt)

Next, let's find the individual derivatives:

d(e−t)/dt = −e−t (since the derivative of e−t is −e−t)

d(t)/dt = 1

Substituting these values back into dy/dt:

dy/dt = t(-e−t) + te−t

Now, we need to find dt/dx. Since x = et, we can solve for dt/dx by differentiating both sides of the equation:

dx/dt = d(et)/dt

Using the chain rule, we have:

dx/dt = (d(et)/d(et)) * (d(et)/dt)

Differentiating et with respect to et gives 1, and differentiating et with respect to t gives et. Therefore:

dx/dt = 1 * et = et

Now, we can solve for dt/dx by taking the reciprocal of dx/dt:

dt/dx = 1/(et) = e−t

Finally, we can compute dy/dx by multiplying dy/dt by dt/dx:

dy/dx = (dy/dt) * (dt/dx)

= (t(-e−t) + te−t) * (e−t)

= -t * e−t + t * e−t

= t(e−t - e−t)

= 0

Therefore, dy/dx = 0.

Now, let's find d2y/dx2:

To find d2y/dx2, we will differentiate dy/dx with respect to x.

Since dy/dx is equal to 0, differentiating it with respect to x will still yield 0.

Therefore, d2y/dx2 = 0.

Given that:

x = et

y = te−t

First, let's compute dy/dx:

To find dy/dx, we will differentiate y with respect to t and then multiply it by dt/dx.

Differentiating y with respect to t:

dy/dt = d(te−t)/dt

Using the product rule, we have:

dy/dt = t(d(e−t)/dt) + (te−t)(d(t)/dt)

Next, let's find the individual derivatives:

d(e−t)/dt = −e−t (since the derivative of e−t is −e−t)

d(t)/dt = 1

Substituting these values back into dy/dt:

dy/dt = t(-e−t) + te−t

Now, we need to find dt/dx. Since x = et, we can solve for dt/dx by differentiating both sides of the equation:

dx/dt = d(et)/dt

Using the chain rule, we have:

dx/dt = (d(et)/d(et)) * (d(et)/dt)

Differentiating et with respect to et gives 1, and differentiating et with respect to t gives et. Therefore:

dx/dt = 1 * et = et

Now, we can solve for dt/dx by taking the reciprocal of dx/dt:

dt/dx = 1/(et) = e−t

Finally, we can compute dy/dx by multiplying dy/dt by dt/dx:

dy/dx = (dy/dt) * (dt/dx)

= (t(-e−t) + te−t) * (e−t)

= -t * e−t + t * e−t

= t(e−t - e−t)

= 0

Therefore, dy/dx = 0.

Now, let's find d2y/dx2:

To find d2y/dx2, we will differentiate dy/dx with respect to x.

Since dy/dx is equal to 0, differentiating it with respect to x will still yield 0.

Therefore, d2y/dx2 = 0.