a. The gradient of [tex]f[/tex] is simply the vector of its partial derivatives:
[tex]\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath=\boxed{2\cos(2x+5y)\,\vec\imath+5\cos(2x+5y)\,\vec\jmath}[/tex]
b. Compute the value of the gradient at [tex]x=-5[/tex] and [tex]y=2[/tex]:
[tex]\nabla f(-5,2)=2\cos0\,\vec\imath+5\cos0\,\vec\jmath=\boxed{2\,\vec\imath+5\,\vec\jmath}[/tex]
c. The directional derivative of [tex]f[/tex] at (-5, 2) in the direction of a vector [tex]\vec u[/tex] is
[tex]\nabla f(-5,2)\cdot\dfrac{\vec u}{\|\vec u\|}[/tex]
It's not entirely clear what the given vector [tex]\vec u[/tex] is supposed to be... I'll consider a more general vector,
[tex]\vec u=a\,\vec\imath+b\,\vec\jmath[/tex]
Then the derivative of [tex]f[/tex] at [tex]P[/tex] in the direction of [tex]\vec u[/tex] is
[tex](2\,\vec\imath+5\,\vec\jmath)\cdot\dfrac{a\,\vec\imath+b\,\vec\jmath}{\sqrt{a^2+b^2}}=\boxed{\dfrac{2a+5b}{\sqrt{a^2+b^2}}}[/tex]
