Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 24

Determine the equation of the tangent line to the curve $\displaystyle y = \frac{1}{\sin x + \cos x}$ at the given point $(\displaystyle 0,1)$


$
\begin{equation}
\begin{aligned}

y' =& \frac{(\sin x + \cos x) \displaystyle \frac{d}{dx} (1) - \left[ \frac{d}{dx} (\sin x + \cos x) \right]}{(\sin x + \cos x)^2}
&& \text{Using Quotient Rule}
\\
\\
y' =& \frac{(\sin x + \cos x) (0) - (\cos x - sin x)}{(\sin x + \cos x)^2}
&&
\\
\\
y' =& \frac{\sin x - \cos x}{(\sin x + \cos x)^2}
&&
\\
\\
& \text{Let $y' = m_T$ (slope of the tangent line)}
&&
\\
\\
y' = m_T =& \frac{\sin (0) - \cos (0)}{[\sin (0) + \cos (0)^2]}
&& \text{Substitute value of $x$}
\\
\\
m_T =& \frac{0 - 1}{(0 + 1)^2}
&&
\\
\\
m_T =& -1

\end{aligned}
\end{equation}
$


Using Point Slope Form substitute the values of $x, y$ and $m_T$


$
\begin{equation}
\begin{aligned}

y - y_1 =& m (x - x_1)
\\
\\
y - 1 =& -1 (x - 0)
\\
\\
y - 1 =& -x
\\
\\
y =& -x + 1
\\
\\
\end{aligned}
\end{equation}
$