Partial derivatives, why is the fy this?

[mr_coffee]

Hello everyone,i had a question..i have the following problem, I'm suppose to find the first partial derivatives:
f(x,y,z,t) = xyz^2*tan(yt);
I got all the partial derivaties right but the fy, they get:
fy = xyz^2*sec^2(yt);
when i do it, i get:
fy = xz^2*sec^2(yt)*t = txz^2*sec^2(yt);
Arn't u suppose to think x,z and t are all constants? if that's the case, when u use the chain rule inside the sec^2(yt), d/dt(yt) = t isn't it?

 

[TD]

Hmm, I actually don't agree with either of those answers.

If all the variables are independent, you have to consider x,z and t as constants when deriving with respect to y, that's correct. But you have to use the product rule as well, there's a y in "xyz^2" and again one in the tangent...

 

[HallsofIvy]

Hello everyone,i had a question..i have the following problem, I'm suppose to find the first partial derivatives:
f(x,y,z,t) = xyz^2*tan(yt);
I got all the partial derivaties right but the fy, they get:
fy = xyz^2*sec^2(yt);

If f really is as you write it, that's clearly wrong.

when i do it, i get:
fy = xz^2*sec^2(yt)*t = txz^2*sec^2(yt);

But that's also clearly wrong!

Arn't u suppose to think x,z and t are all constants? if that's the case, when u use the chain rule inside the sec^2(yt), d/dt(yt) = t isn't it?

Yes, doing a partial derivative, you treat all other variables as constants. But in addition to using the chain rule, you also have to use the product rule!

f_y= xz²tan(yt)+ txyz²sec²(yt)