- #1
Yankel
- 395
- 0
Hello,
I have a tricky chain rule question, I think understanding it is more difficult than solving.
For the function z=f(x,y) it is given that:
f_{y}(0,-3)=-2
and
\[f_{x}(0,-3)=3\]
so for the function
\[g(x,y)=f(2\cdot ln(x+y),x^{4}-3y^{2})\]
choose the correct answer:
(1)
\[g_{y}(0,-3)=18\]
(2)
\[g_{y}(0,-3)=-6\]
(3)
\[g_{y}(0,1)=18\]
(4)
\[g_{y}(0,1)=-6\]
(5)
Non of the above answers
I am confused slightly. I thought to call
\[u=2\cdot ln(x+y)\]
and
\[v=x^{4}-3y^{2}\]
but when I put x=0 and y=-3, I can't get a value for u, since I get a negative value under the ln.
Can you assist clearing this up ? Thank you !
I have a tricky chain rule question, I think understanding it is more difficult than solving.
For the function z=f(x,y) it is given that:
f_{y}(0,-3)=-2
and
\[f_{x}(0,-3)=3\]
so for the function
\[g(x,y)=f(2\cdot ln(x+y),x^{4}-3y^{2})\]
choose the correct answer:
(1)
\[g_{y}(0,-3)=18\]
(2)
\[g_{y}(0,-3)=-6\]
(3)
\[g_{y}(0,1)=18\]
(4)
\[g_{y}(0,1)=-6\]
(5)
Non of the above answers
I am confused slightly. I thought to call
\[u=2\cdot ln(x+y)\]
and
\[v=x^{4}-3y^{2}\]
but when I put x=0 and y=-3, I can't get a value for u, since I get a negative value under the ln.
Can you assist clearing this up ? Thank you !