First, find $\displaystyle f_{x} = \frac{\partial f}{\partial x}$ as follows:

$\displaystyle f (x,y) = e ^{x^{2}+ y^{2}} = e ^{x^{2} } e ^{y^{2} }$

$\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \left (e ^{x^{2} } e ^{y^{2} } \right )$

$\displaystyle = e ^{y^{2}} \cdot \frac{\partial }{\partial x} \left (e ^{x^{2} }} \right )$

$\displaystyle = e ^{y^{2}} \cdot \frac{\partial }{\partial x} x^{2} \cdot \left (e ^{x^{2} }} \right )$

$\displaystyle = 2x \cdot e ^{y^{2}} \cdot \left (e ^{x^{2} }} \right )$

$\displaystyle = 2x e ^{x^{2}} e ^{y^{2}}$

Therefore, $\displaystyle f_{x} (x,y ) = 2x e ^{x^{2}} e ^{y^{2}}$.

Now, find $\displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$\displaystyle f_{x} (x,y )= 2x e ^{x^{2}} e ^{y^{2}}$

$\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y} \left ( 2x e ^{x^{2}} e ^{y^{2}} \right )$

$\displaystyle =2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( e ^{y^{2}} \right )$

$\displaystyle =2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( y^{2} \right ) e ^{y^{2}}$

$\displaystyle =2x e ^{x^{2}} \cdot 2y \cdot e ^{y^{2}}$

$\displaystyle =4xy e ^{x^{2}} e ^{y^{2}} = 4xy e ^{x^{2}+y^{2}}$

First, find $\displaystyle f_{x} = \frac{\partial f}{\partial x}$ as follows:

$\displaystyle f (x,y) = \cos (x^{2} + y^{2})$

$\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \cos (x^{2} + y^{2})$

$\displaystyle = \frac{\partial }{\partial x}(x^{2} + y^{2}) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]$

$\displaystyle = (2x + 0) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]$

$\displaystyle = - 2x\sin (x^{2} + y^{2})$

Therefore, $\displaystyle f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2})$.

Now, find $\displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$\displaystyle f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2})$

$\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left [ - 2x\sin (x^{2} + y^{2}) \right ]$

$\displaystyle =- 2x \frac{\partial }{\partial y}\left [ \sin (x^{2} + y^{2}) \right ]$

$\displaystyle =- 2x\cdot \frac{\partial }{\partial y}(x^{2} + y^{2}) \cdot \cos (x^{2} + y^{2})$

$\displaystyle =- 2x\cdot (0 + 2y) \cdot \cos (x^{2} + y^{2})$

$\displaystyle =- 2x\cdot ( 2y) \cdot \cos (x^{2} + y^{2})$

$\displaystyle =- 4xy \cos (x^{2} + y^{2})$

First, find $\displaystyle f_{x} = \frac{\partial f}{\partial x}$ as follows:

$\displaystyle f (x,y) = \log xy = \log x + \log y = \frac{ \ln x }{\ln 10}+ \frac{ \ln y }{\ln 10}$

$\displaystyle f (x,y) = \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right )$

$\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x}\left [ \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right ) \right ]$

$\displaystyle = \frac{ 1}{\ln 10} \left (\frac{\partial }{\partial x}\left \ln x+\frac{\partial }{\partial x} \ln y \right )$

$\displaystyle = \frac{ 1}{\ln 10} \left ( \frac{1}{x}+0 \right )$

$\displaystyle = \frac{ 1}{\ln 10} \cdot \frac{1}{x}$

Therefore, $\displaystyle f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}$.

Now, find $\displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$\displaystyle f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}$

$\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left ( \frac{ 1}{\ln 10} \cdot \frac{1}{x} \right )$

$\displaystyle =0$

Rewrite the equation to be solved for $\displaystyle x$ as $\displaystyle e^{x} - \sin x = 0$.

Let $\displaystyle f(x ) = e^{x} - \sin x$. 

$\displaystyle f'(x ) = \frac{\mathrm{d} }{\mathrm{d} x} \left (e^{x} - \sin x \right )$

$\displaystyle f'(x ) = e^{x} - \cos x \right )$

The problem amounts to finding a zero of $\displaystyle f(x )$. By Newton's method, the second approximation can be derived from the first using the equation

$\displaystyle x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$.

Since $\displaystyle x_{1} = -2$, 

$\displaystyle f(-2 ) = e^{-2} - \sin (-2) \approx 0.1353 - (-0.9093 ) \approx 1.0446$ 

and

$\displaystyle f'(-2 ) = e^{-2} - \cos (-2) \approx 0.1353 - \left ( -0.4161 \right )\approx 0.5514$

Use these to find the approximation:

$\displaystyle x_{2} =-2 - \frac{f(-2)}{f'(-2)} \approx -2 - \frac{1.0446}{0.5514}\approx -2 -1.8945 \approx -3.895$

This function is continuous and differentiable everywhere. First we find the value(s) of $\displaystyle x$ for which $\displaystyle f'(x) =0$.

$\displaystyle f(x) = e^x- 4x$

$\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left ( e^x- 4x \right )$

$\displaystyle f'(x) = e^x- 4$

$\displaystyle f'(x) =0$

$\displaystyle e^x- 4 =0$

$\displaystyle e^x= 4$

$\displaystyle \ln e^x= \ln 4$

$\displaystyle x= \ln 4$

Therefore, this is the only possible minimum. We determine whether it is a minimum by evaluating $\displaystyle f''(\ln 4)$:

$\displaystyle f'(x) = e^x- 4$

$\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^x- 4 \right )$

$\displaystyle f''(x) = e^x$

$\displaystyle f''(\ln 4) = e^{\ln 4} = 4 > 0$

Since $\displaystyle f''(\ln 4) > 0$, $\displaystyle f$ has its minimum value at $\displaystyle x= \ln 4$; it is

$\displaystyle f( \ln 4) = e^{ \ln 4}- 4 \cdot \ln 4 = 4 - 4 \ln 4$.

$\displaystyle y =\ln \left ( \sin x + \cos y \right )$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \sin x + \cos y \right )$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )}{\sin x + \cos y}$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \left ( \sin x + \cos y \right )= \frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \sin x + \frac{\mathrm{d} y }{\mathrm{d} x} \cos y \right )= \cos x - \frac{\mathrm{d}y }{\mathrm{d} x} \sin y$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \sin x + \frac{\mathrm{d}y }{\mathrm{d} x} \sin y+ \frac{\mathrm{d} y }{\mathrm{d} x} \cos y \right )= \cos x$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \left (\sin x + \sin y+ \cos y \right )= \cos x$

$\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y+ \cos y }$

$\displaystyle f(x) = 4x + 8 \sin x$

$\displaystyle f'(x) = 4 + 8 \cos x$

We first look for $\displaystyle x$ such that $\displaystyle f'(x) = 0$:

$\displaystyle 4 + 8 \cos x = 0$

$\displaystyle 8 \cos x = -4$

$\displaystyle \cos x = -\frac{1}{2}$

The two values on the interval $\displaystyle [-\pi, \pi]$ for which this holds true are $\displaystyle -\frac{2}{3} \pi , \frac{2}{3} \pi$, so we evaluate $\displaystyle f$ for the values $\displaystyle - \pi, -\frac{2}{3} \pi , \frac{2}{3} \pi, \pi$:

$\displaystyle f(-\pi ) = 4(-\pi ) + 8 \sin (-\pi ) = -4 \pi + 8 \cdot 0 = -4 \pi \approx -12. 5$

$\displaystyle f \left \(-\frac{2}{3} \pi\right$ = 4\left $-\frac{2}{3} \pi\right$ + 8 \sin \left $-\frac{2}{3} \pi\right$\)

     $\displaystyle = -\frac{8}{3} \pi + 8 \cdot \left (- \frac{\sqrt{3}}{2} \right ) = -\frac{8}{3} \pi - 4 \sqrt {3} \approx -15.3$

$\displaystyle f \left \(\frac{2}{3} \pi\right$ = 4\left $\frac{2}{3} \pi\right$ + 8 \sin \left $\frac{2}{3} \pi\right$\)

     $\displaystyle = \frac{8}{3} \pi + 8 \cdot \left ( \frac{\sqrt{3}}{2} \right ) = \frac{8}{3} \pi + 4 \sqrt {3} \approx 15.3$

$\displaystyle f(\pi ) = 4\pi + 8 \sin \pi = 4 \pi + 8 \cdot 0 = 4 \pi \approx 12. 5$

The minimum value is $\displaystyle f \left \(-\frac{2}{3} \pi\right$ = -\frac{8}{3} \pi - 4 \sqrt {3}\).

$\displaystyle x^2 + y^2 = e^ { xy }$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^2 + y^2 \right )= \frac{\mathrm{d} }{\mathrm{d} x}e^ { xy }$

$\displaystyle 2x + 2y \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} xy \cdot e^ { xy }$

$\displaystyle 2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \left ( x \cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y \right ) \cdot e^ { xy }$

$\displaystyle 2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y e^ { xy }$

$\displaystyle 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} - x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} =-2x+ y e^ { xy }$

$\displaystyle 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} - x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} =-2x+ y e^ { xy }$

$\displaystyle \left (2y- x e^ { xy } \right )\cdot \frac{\mathrm{d} y}{\mathrm{d} x} =-2x+ y e^ { xy }$

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}$

First, we determine if there are any points at which $\displaystyle f'(x) = 0$.

$\displaystyle f(x) = \frac{x}{2}+ \sin x = \frac{1}{2}x+ \sin x$

$\displaystyle f'(x) = \frac{1}{2} + \cos x$

$\displaystyle f'(x) = 0$

$\displaystyle \frac{1}{2} + \cos x = 0$

$\displaystyle \cos x = -\frac{1}{2}$

The only point on the interval on which this is true is $\displaystyle x =\frac{ 2\pi }{3}$.

We test this point as well as the two endpoints, $\displaystyle x = 0$ and $\displaystyle x = \pi$, by evaluating $\displaystyle f$ for each of these values.

$\displaystyle f(0) = \frac{0}{2}+ \sin 0 = 0 + 0 = 0$

$\displaystyle f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \left ( \frac{ 2\pi }{3} \right )}{2}+ \sin \frac{ 2\pi }{3} = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} \approx 2$

$\displaystyle f( \pi ) = \frac{ \pi}{2}+ \sin \pi = \frac{ \pi}{2} + 0 = \frac{ \pi}{2} \approx 1.6$

Therefore, $\displaystyle f$ assumes its maximum on this interval at the point $\displaystyle x =\frac{ 2\pi }{3}$, and $\displaystyle f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} = \frac{ 2\pi + 3\sqrt{3} }{6}$.

$\displaystyle \sec y = x^2 + xy^2$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec y =\frac{\mathrm{d} }{\mathrm{d} x} \left (x^2 + xy^2 \right )$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + x \cdot 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + 2xy \frac{\mathrm{d}y }{\mathrm{d} x}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y - \frac{\mathrm{d}y }{\mathrm{d} x} 2xy=2x + y^2$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}$