1

z=f(x,y) , x= s^2+3t, y= s+2t^2 (∂z/∂t)= 3fx+4tfy why does (∂/∂t)(3fx+4tfy) = 3(∂/∂t)fx + 4fy + 4t(∂/∂t)fy ? I don't understand where 4fy comes from. Say it would have been (∂/∂t)(3t^2fx+4tfy) is this equal to 3t^2(∂/∂t)fx +6tfx + 4fy + 4t(∂/∂t)fy ?

enter image description here

John
  • 33
  • No I don't. Is the statement I wrote correct? – John Oct 29 '13 at 02:37
  • Yes it is, not sure I understand it completely but I'm more confused where the term 4fy comes from. I can see that it's the derivative of 4tfy, but I don't understand why you do it. – John Oct 29 '13 at 02:43
  • so the partial of 4tfy wrt t = the partial of 4tfy wrt t + partial of 4tfy wrt fy? I've tried to solve (∂^2z/∂s^2)= (∂/∂s)*(∂z/∂s)=(∂/∂s)(2sfx+fy)=2s(∂/∂s)fx+2fx+(∂/∂s)fy=2s(2sfxx+fxy)+2fx+2sfxy+fyy=4s^2fxx+2fx+4sfxy+fyy. Is this correct or have I misunderstood it? – John Oct 29 '13 at 02:51
  • Thanks a lot! I think I got it now, could you just check if the comment I edited is correct? Or did I make some mistake in it? – John Oct 29 '13 at 03:03
  • Can you follow what you posted? – Amzoti Oct 29 '13 at 03:37
  • I think so yeah, I did (∂^2z/∂s^2) the same way in the comment. Thanks a lot! – John Oct 29 '13 at 05:29

1 Answers1

3

We want to take the partial wrt $t$ of the expression:

$$4 \times t \times f_y$$

So, we write (using the product rule):

$\dfrac{\partial}{\partial t}( 4~ t ~ f_y) = 4 \dfrac{\partial}{\partial t}( t \times f_y) = 4 \left(\dfrac{\partial}{\partial t}(t)\times f_y + t \times \dfrac{\partial}{\partial t}(f_y)\right) = 4 \left(f_y + t\dfrac{\partial}{\partial t}(f_y)\right)$

Amzoti
  • 56,093