1

The question states given that $f(x,y)$ has gradient $f = (2xy,x^2),x(s,t)=s+t,y(s,t) = st$ and $g(s,t) = f(x(s,t),y(s,t))$ calculate $\frac{\partial g}{\partial s}(1,2)$ so this is what my friend did $$\frac{\partial g}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=2xy\cdot 1+x^2\cdot t=2\cdot1\cdot2\cdot1+1^2\cdot2=4+2=6$$

These are now my thoughts, $x(s,t) = 1+2 = 3$ and $y(s,t)=1(2)=2$ so doing something very similar to him I computed $$\frac{\partial g}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=2xy\cdot1+x^2\cdot t = 2\cdot3\cdot2\cdot1+3^2\cdot2=30$$

If anyone could inform me of who is correct and WHY that would be greatly appreciated!

overlord12
  • 167

1 Answers1

1

hint

From the given gradient of $ f $, we derive $$f(x,y)=x^2y+C$$ and

$$g(s,t)=(s+t)^2st+C$$ $$=s^3t+2s^2t^2+st^3+C$$

thus

$$\frac{\partial g}{\partial s}(s,t)=3s^2t+4st^2+t^3$$ it gives $30$.

hamam_Abdallah
  • 62,951