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Answer:
Remember that the gradient is equivalent o the derivative evaluated in one point.
And remember that if we have a function like:
f(x) a*x^n
The derivative is:
f'(x) = n*a*x^(n - 1)
Then knowing this, let's start:
a) y = x^2
the derivative is:
y' = 2*x
And we want to evaluate this in the point (3, 9), then we just need to replace x by 3.
y'(3) = 2*3 = 6
The gradient of y = x^2 at (3, 9) is 6.
b) y = -4*x^3
The derivative is:
y' = 3*(-4*x^2) = -12*x^2
And we want to evaluate this on (1, -4), then we need to replace x by 1.
y'(1) = -12*1^2 = -12
The gradient of y = -4*x^3 at (1, -4) is -12.
c) y = 2*x^4
The derivative is:
y' = 4*2*(x^3) = 8*x^3
And we want to evaluate this on (-1, 2), then we need to replace x by -1.
y'(-1) = 8*(-1)^3 = -8
The gradient of y = 2*x^4 at the point (-1, 2) is -8
d) y = 6*x
The derivative is:
y' = 6
We can see that here we do not have any dependence on x, so the gradient will be the same for all values of x, this means that the gradient at the point (2, 12) is 6.