In a Calculus 2-exam, the candidates were presented with the following problem:
a) Let $f(x,y,z) = \sin^2(2x)\ln(5y+z^2)$. Compute the gradient.
b) Find the directional derivative in the direction u = $(1,2,3)$ at $(x,y,z) = (\pi/4, 1, 0)$
What follows is what I would have replied on an exam. I ask you to find eventual errors, comment on clarity, and if possible, score the answer on a scale from $0$ to $6$:
a) The gradient $\nabla f$ is defined as $\nabla f = f_1 \hat{i} + f_2\hat{j} + f_3 \hat{k}$. First, we find all the partial derivatives:
$f_1 = 2\sin(2x)2\cos(2x)\ln(5y+z^2)\\ f_2=\sin^2(2x)\cdot\frac{5}{5y+z^2}\\ f_3 = \sin^2(2x)\cdot\frac{2z}{5y+z^2}$
The gradient is now given by $\nabla f = f_1\hat{i} + f_2\hat{j} + f_3\hat{k}$, as stated above.
b) We first normalize the vector u. We call this vector v. Then v = $(1/\sqrt{14},2/\sqrt{14},3/\sqrt{14})$. Then we may compute our directional derivative as $\nabla f(\pi/4, 1, 0) \bullet$ v = $2/\sqrt{14}$.
Your $f_1$ is missing a factor of $2$ from the chain-rule for the inside function $2x$. Beyond that, it looks good generally, I might suppose you are suppose to evaluate the expressions at the given point so given that lack of detail (which many students will mess-up the details so to be fair I have to suppose you might...) I'd say 3/6. Put in that detail 5/6 given that you appear to know the chain rule and the general structure. Perhaps 5.5 would be fair given that the multivariable concepts are on target.
EDIT: upon a second more careful read, 6/6.