(a)
\begin{align*}
f(x,y)=x^{2}+y^{2}, \quad x=\cos t + \sin t, \quad y=2t, \quad \text{find} \quad \frac{d f}{d t}
\end{align*}
Solution (a):
\begin{align*}
&f'_{t} = 2x x'_{t} + 2y y'_{t} = 2(\cos t + \sin t) (-\sin t +\cos t) + 2\cdot 2t\cdot2=\\
&2(\cos^{2}t-\sin^{2}t)+8t
\end{align*}
(b)
\begin{align*}
z=4e^{x}\ln y, \quad x = \ln(u \cos v), \quad y=u\sin v \quad \text{find} \quad \frac{\partial z}{\partial u}
\end{align*}
Solution (b):
\begin{align*}
&z'_{u}=4e^{x}x'_{u} \cdot \ln y + 4e^{x}\cdot \frac{1}{y}y'_{u}=\\
&4e^{\ln u\cos v}\cdot \frac{1}{u} \cdot \ln (u \sin v) + 4e^{\ln u
\cos v}\cdot \frac{1}{u\sin v } \cdot \sin v=\\
&4\cos v \ln(u\sin v) + 4\cos v
\end{align*}
Let $f(x,y)=2x^{2}+y^{2}.$ Find the derivative of the function at the point
$(-1,1)$ in the direction: $\vec{v} = 3\vec{i}-4\vec{j}$.
Solution:
Derivative of this function along this direction by definition is
$$
\nabla f (-1,1) \cdot \frac{\vec{v}}{|\vec{v}|}=\langle f'_{x}(-1,1),f'_{y}(-1,1)\rangle \cdot \frac{\vec{v}}{|\vec{v}|}
$$
\begin{align*}
f'_{x} &= 4x, \quad \text{so} \quad f'_{x}(-1,1)=-4. \\
f'_{y} &=2y, \quad \text{so} \quad f'_{y}(-1,1)=2,\\
\frac{\vec{v}}{|\vec{v}|} &= \frac{3\vec{i}-4\vec{j}}{\sqrt{3^{2}+(-4)^{2}}}=\\
\frac{3}{5}\vec{i} &- \frac{4}{5}\vec{j} = \langle\frac{3}{5}, -\frac{4}{5} \rangle.\\
\end{align*}
Thus
\begin{align*}
&\langle f'_{x}(-1,1),f'_{y}(-1,1)\rangle \cdot \frac{\vec{v}}{|\vec{v}|}=
\langle -4,2\rangle\cdot \langle\frac{3}{5}, -\frac{4}{5} \rangle = \\
&\frac{-12}{5}-\frac{8}{5}=-\frac{20}{5}=-4.
\end{align*}