Find the first partial derivatives of the function: f(x,y,z,t) = xyz^2tan(yt)

Question

anonymous · Answer

$$f(x,y,z,t) = xyz^2\tan(yt) $$

unklerhaukus · Answer

$$\frac{\partial}{\partial x} f(x,y,z,t)\
= \frac{\partial}{\partial x}xyz^2\tan(yt)\
= \frac{\partial}{\partial x}\big(x\big)\cdot yz^2\tan(yt)+x\cdot \frac{\partial}{\partial x}\big(yz^2\tan(yt)\big)\=\=$$

anonymous · Answer

Partial Derivative in respect to x is:
$$yz^2(1)\tan(yt) => yz^2\tan(yt)$$

unklerhaukus · Answer

yes!

anonymous · Answer

Cool thanks
Now I just have to do the rest of the partial derivatives for y, z and t now. But I'll think I'll take it form here!

unklerhaukus · Answer

the partial with respect to y is a bit trickier,

anonymous · Answer

Oh yeah, possibly product rule ~

unklerhaukus · Answer

yeah, and the chain rule too

anonymous · Answer

And that too, for that trig function part