Partial derivative with chain rule: check work

[939]

Homework Statement

If possible, please check my work for any large errors.

y = 10kl - √k - √l
k = (t/5) + 5
l = 5e^t/10

Evaluate at t = 0 using chain rule.

Homework Equations

y = 10kl - √k - √l
k = (t/5) + 5
l = 5e^t/10

The Attempt at a Solution

= ∂y/∂k * dk/dt + ∂y/∂l * dl/dt
= (10l - 0.5K^-0.5)(1/5) + (10k - 0.5l^-.05)((e^(x/10))/2)

k(0) = 5
l(0) = 5
y(0) = ~9.95 + ~41.03 = ~51

 

[Mark44]

Homework Statement

If possible, please check my work for any large errors.

y = 10kl - √k - √l
k = (t/5) + 5
l = 5e^t/10

Evaluate at t = 0 using chain rule.

Homework Equations

y = 10kl - √k - √l
k = (t/5) + 5
l = 5e^t/10

The Attempt at a Solution

= ∂y/∂k * dk/dt + ∂y/∂l * dl/dt
= (10l - 0.5K^-0.5)(1/5) + (10k - 0.5l^-.05)((e^(x/10))/2)

The above is close. The -0.5l^(-.05) term should be -0.5l^(-0.5). I don't know if that was a transcription error. Also, the exponential term would be clearer as (1/2)e^(x/10).

k(0) = 5
l(0) = 5
y(0) = ~9.95 + ~41.03 = ~51

k(0) and l(0) are fine, but I didn't check your other numbers.

One other thing. It's not good to start your first line with "=". For your problem, you should start with dy/dt = ...