- #1
iScience
- 466
- 5
could someone show me how [itex]\frac{∂}{∂t}[/itex]([itex]\frac{∂z}{∂u}[/itex])= [itex]\frac{∂^2z}{∂u^2}[/itex] [itex]\frac{∂u}{∂t}[/itex]
where z=z(u)
u=x+at
where z=z(u)
u=x+at
iScience said:could someone show me how [itex]\frac{∂}{∂t}[/itex]([itex]\frac{∂z}{∂u}[/itex])= [itex]\frac{∂^2z}{∂u^2}[/itex] [itex]\frac{∂u}{∂t}[/itex]
where z=z(u)
u=x+at
iScience said:i apologize, but can you please show me what you did step by step? i know how to apply the chain rule for functions when the function itself is actually given, ie the derivative of the function times the derivation of the inside. but i don't understand what we're doing now.
where did you get the d/du in the front from? and the ∂u/∂t that's behind z', that's just from the derivative of the inside right?
Your problem is solved by applying one of these chain rules to z′(u) with second variables x and t instead of x and y.
iScience said:i understand z'(t)=[itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂t}[/itex] and z'(x)=[itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂x}[/itex]
i just don't understand the leap from that to this: [itex]\frac{∂}{∂t}[/itex][itex]\frac{∂z}{∂u}[/itex]
but 'u' is the intermediate variable. i don't know what to do with [itex]\frac{∂z}{∂u}[/itex]. if it involved finding the partial of z with respect to x and t, i could just apply the chain rule like you say, but it's asking me to find the partial of z with respect to u, the intermediate variable. what can i do with that? i can't use x and y.
iScience said:okay can you tell me if i did this correctly
[itex]\frac{∂}{∂t}[/itex]([itex]\frac{∂z}{∂u}[/itex])=[itex]\frac{∂}{∂u}[/itex]([itex]\frac{∂z}{∂t}[/itex])=[itex]\frac{∂}{∂u}[/itex]([itex]\frac{∂z}{∂u}[/itex][itex]\frac{∂u}{∂t}[/itex])
i basically just swapped the positions of the u and t in the beginning is this right?
if i swap the u and the t, aren't i applying clairaut's theorem? and if i am, then don't i need to prove that zut is continuous and ztu is continuous since that's the condition for which clairaut's theorem applies?
The partial derivative chain rule is a mathematical tool used to calculate the derivatives of multivariable functions. It helps us find the rate of change of a function with respect to one variable while holding all other variables constant.
The partial derivative chain rule is used for functions with multiple variables, while the regular chain rule is used for functions with a single variable. The partial derivative rule involves taking derivatives with respect to one variable at a time, while the regular chain rule involves taking the derivative of a composite function.
Sure, let's say we have a function f(x,y) = x^2 + 3y - 5. To find the partial derivative of this function with respect to x, we would use the partial derivative chain rule and treat y as a constant. This would give us the result f'(x,y) = 2x.
One common mistake is forgetting to treat the other variables as constants when taking the derivative with respect to a specific variable. It's also important to carefully apply the chain rule when dealing with nested functions. Another mistake is mixing up the order of the variables when taking partial derivatives.
The partial derivative chain rule is commonly used in fields such as physics, economics, and engineering to find rates of change in multivariable systems. For example, it can be used to calculate the marginal cost in economics or the acceleration of an object in physics.