
has gradient

which at the point (-1, 4, 3) has a value of

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say
, in which case we have

Then the derivative of
at (-1, 4, 3) in the direction of
is

Answer:
y>-5-3x
Step-by-step explanation:
The bases were not the same when the exponents were set equal to each other.
16 should have been written as 4 squared.
The exponent on the right should be 4a instead of 8a.
The correct solution is a=-3/4.
Answer:
9
Step-by-step explanation:
10-((-3) + -6)-10
10-(-9)-10
10+9-10
9