2. \( 2 x d x-2 y d y=x^{2} y d y-2 x y^{2} d x \) 3. \( x \sqrt{1+y^{2}}+y y^{\prime} \sqrt{1+x^{2}}=0 \) 4. \( 4 x d x-3 y d y=3 x^{2} y d y-2 x y^{2} d x \)

implications. 10. Deseribe how vector calculas is utilized in electromagnetiom, specif- eally in calculating electric flelds.

\( \left. \begin{array} { l l } { u = x ^ { 2 } + 3 x + 2 } & { u ^ { \prime } = 2 x + 3 } \\ { v = x - 1 } & { v ^ { 3 } = 1 } \\ { } & { ( 2 x + 3 ) ( x - 1 ) - ( x ^ { 2 } + 3 x ) 2 ) ( 1 ) } \end{array} \right. \)

\( \left\{\begin{array}{l}f(x)=x^{2}-1 \\a=1 \\\lim _{h \rightarrow 0} (\frac{f(a+h-f(a)}{h}\end{array}\right. \)

\( y = ( 4 x ^ { 2 } - 2 x + 6 ) ( x ^ { 2 } + 6 x - 3 ) \quad y ^ { \prime } = \)

\( \left. \begin{array} { l } { u = x ^ { 2 } + 3 x + 2 \quad u ^ { \prime } = 2 x + 3 } \\ { V = x - 1 \quad V ^ { \prime } = 1 } \\ { \frac { ( 2 x + 3 ) ( x - 7 ) - ( x ^ { 2 } + 3 x + 2 ) ( 1 ) } { ( x - 1 ) ^ { 2 } } } \end{array} \right. \)