Partial Derivative Homework: Calculate ∂f/∂x

[dyn]

Homework Statement

The question asks to calculate ∂f/∂x for f(x,y,t) = 3x² + 2xy + y^1/2t -5xt where x(t) = t³ and y(t) = 2t⁵

Homework Equations

The answer is given as ∂f/∂x = 6x + 2y - 5t

The Attempt at a Solution

I'm confused because the answer given seems to treat x,y ,t as independent variables and the answer given is just a partial derivative treating y and t as constant. But really x,y.t are all dependent on each other. Is it even possible to obtain a partial derivative with respect to x in this case ?

 

[andrewkirk]

The information that x(t) = t³ and y(t) = 2t⁵ is a red herring. When taking a partial derivative of f with respect to x, we look only at the explicit, direct role of x in the formula for f(x,y,t). We ignore any dependencies, and that leads to the quoted result. If we want to take dependencies into account, we take a total derivative, which is written $\frac{df}{dx}$, rather than the partial derivative $\frac{\partial f}{\partial x}$.

The Insight article on partial derivatives gives more background on partial derivatives, and the nature of these distinctions.

 

[dyn]

Thanks. And would I be correct that the total derivative is
df/dx = ∂f/∂t + (∂f/∂x)dx/dt + (∂f/∂y)dy/dt ?

 

[andrewkirk]

Thanks. And would I be correct that the total derivative is
df/dx = ∂f/∂t + (∂f/∂x)dx/dt + (∂f/∂y)dy/dt ?

That's the formula for the total derivative with respect to t.
The total derivative wrt x uses the same formula, but swaps the role of t and x on the RHS, giving
$$\frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t} \frac{d t}{dx}
+ \frac{\partial f}{\partial y} \frac{dy}{dx}$$

 

[dyn]

Thanks for your help