you turn 1 and 1 over 5 into 6 over 5. then divide.
12 divided by 6 over 5 equals 10
The answer to this problem is 4

a. The gradient is


b. The gradient at point P(1, 2) is

c. The derivative of
at P in the direction of
is

It looks like

so that

Then


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Answer:56
Step-by-step explanation:
i got it correct on my test