Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 24

Show that the formulas for the derivatives of the functions a.) $\cos hx$, b.) $\tan hx$, c.) $\csc hx$ ,d.) $\sec hx$ and e.) $\cot hx$

a.) $\cos hx$


$
\begin{equation}
\begin{aligned}

\cos hx =& \frac{e^x + e^{-x}}{2}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{d}{dx} \left( \frac{e^x + e^{-x}}{2} \right)
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{\displaystyle (2) \frac{d}{dx} (e^x + e^{-x}) - (e^x + e^{-x}) \frac{d}{dx} (2) }{(2)^2}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{2 [e^x + (-e^{-x})]}{4}
\\
\\
\frac{d}{dx} (\cos hx) =& \frac{e^x - e^{-x}}{2}

\end{aligned}
\end{equation}
$


We know that $\displaystyle \sin h(x) = \frac{e^x - e^{-x}}{2}$, so

$\displaystyle \frac{d}{dx} (\cos hx) = \sin hx$

b.) $\tan hx$


$
\begin{equation}
\begin{aligned}

\tan hx =& \frac{\sin hx}{\cos hx}
\\
\\
\frac{d}{dx} \tan hx =& \frac{d}{dx} \left( \frac{\sin hx}{\cos hx} \right)
\\
\\
\frac{d}{dx} \tan hx =& \frac{\displaystyle (\cos hx) \frac{d}{dx} (\sin hx) - (\sin hx) \frac{d}{dx} (\cos hx) }{(\cos hx)^2}
\\
\\
\frac{d}{dx} \tan hx =& \frac{(\cos hx) (\cos hx) - (\sin hx)(\sin hx)}{\cos h^2 x}
\\
\\
\frac{d}{dx} \tan hx =& \frac{\cos h^2 x - \sin h^2 x}{\cos h^2 x}


\end{aligned}
\end{equation}
$



We know that $\cos h^2 x - \sin h^2 x = 1$, so


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} \tan hx =& \frac{1}{\cos h^2 x}
\\
\\
\frac{d}{dx} \tan hx =& \sec h^2 x

\end{aligned}
\end{equation}
$


c.) $\csc hx$


$
\begin{equation}
\begin{aligned}

\csc hx =& \frac{1}{\sin hx}
\\
\\
\csc hx =& (\sin hx)^{-1}
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{d}{dx} (\sin hx)^{-1}
\\
\\
\frac{d}{dx} (\csc hx) =& -(\sin hx)^{-2} \frac{d}{dx} (\sin hx)
\\
\\
\frac{d}{dx} (\csc hx) =& - (\sin hx)^{-2} (\cos hx)
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{-1}{\sin h^2 x} \cdot \cos hx
\\
\\
\frac{d}{dx} (\csc hx) =& \frac{- \cos hx}{\sin hx} \cdot \frac{1}{\sin hx}
\\
\\
\frac{d}{dx} (\csc hx) =& - \cot hx \csc hx


\end{aligned}
\end{equation}
$


d.) $\sec hx$


$
\begin{equation}
\begin{aligned}

\sec hx =& \frac{1}{\cos hx}
\\
\\
\sec hx =& (\cos hx)^{-1}
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{d}{dx} (\cos hx)^{-1}
\\
\\
\frac{d}{dx} (\sec hx) =& - (\cos hx)^{-2} \frac{d}{dx} (\cos hx)
\\
\\
\frac{d}{dx} (\sec hx) =& - (\cos hx)^{-2} (\sin hx)
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{-1}{\cos h^2 x} \cdot \sin hx
\\
\\
\frac{d}{dx} (\sec hx) =& \frac{- \sin hx}{\cos hx} \cdot \frac{1}{\cos hx}
\\
\\
\frac{d}{dx} (\sec hx) =& - \tan hx \sec hx

\end{aligned}
\end{equation}
$


e.) $\cot hx$


$
\begin{equation}
\begin{aligned}

\cot hx =& \frac{\csc hx}{\sin hx}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{d}{dx} \left( \frac{\cos hx}{\sin hx} \right)
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{\displaystyle (\sin hx) \frac{d}{dx} (\cos hx) - (\cos hx) \frac{d}{dx} (\sin hx)}{(\sin hx)^2}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{(\sin hx)(\sin hx) - (\cos hx)(\cos hx)}{\sin h^2x}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{\sin h^2 x - \cos h^2 x}{\sin h^2 x}
\\
\\
\frac{d}{dx} (\cot hx) =& \frac{- (\cos h^2 x - \sin h^2 x)}{\sin h^2 x}

\end{aligned}
\end{equation}
$


We know that $\cos h^2 x - \sin h^2 x = 1$


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\cot hx) =& \frac{-1}{\sin h^2 x}
\\
\\
\frac{d}{dx} (\cot hx) =& - \csc h^2 x

\end{aligned}
\end{equation}
$

Comments

Post a Comment

Popular posts from this blog

In “Fahrenheit 451,” what does Faber mean by “Those who don’t build must burn. It’s as old as history and juvenile delinquents”?

During a conversation between Montag and Faber, Faber says, "Those who don't build must burn. It's as old as history and juvenile delinquents" (Bradbury, 42). Faber is saying that those individuals who do not positively contribute to society are helping to destroy it. Faber's analogy corresponds to the function of the firefighters, who destroy books and prevent individuals from attaining knowledge. Montag, who is a jaded firefighter, feels extremely guilty about participating in the destruction of society. He visits Faber in hopes of altering the trajectory of his life and wishes to contribute to society instead of being a destructive force. Faber believes that authors, artists, and intellectuals "build" institutions and have an overall positive impact on society. According to Faber's views, individuals who do not give back to society either directly or indirectly contribute to its destruction, which explains his own feelings of guilt. While Faber do...
Read more

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 34

Determine $y''$ of $\sqrt{x} + \sqrt{y} = 1$ by using implicit differentiation. Solving for 1st Derivative $ \begin{equation} \begin{aligned} \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} (\sqrt{y}) =& \frac{d}{dx} (1) \\ \\ \frac{d}{dx} (x)^{\frac{1}{2}} + \frac{d}{dx} (y)^{\frac{1}{2}} =& \frac{d}{dx} (1) \\ \\ \frac{1}{2} (x)^{\frac{-1}{2}} + \frac{1}{2} (y)^{\frac{-1}{2}} \frac{d}{dx} =& 0 \\ \\ \frac{1}{2 (y)^{\frac{1}{2}}} \frac{dy}{dx} =& \frac{-1}{2(x)^{\frac{1}{2}}} \\ \\ \frac{dy}{dx} =& - \frac{\cancel{2} (y)^{\frac{1}{2}}}{\cancel{2}(x)^{\frac{1}{2}}} \\ \\ \frac{dy}{dx} =& - \frac{(y)^{\frac{1}{2}}}{(x)^{\frac{1}{2}}} \end{aligned} \end{equation} $ Solving for 2nd Derivative $ \begin{equation} \begin{aligned} \frac{d^2 y}{dx^2} =& - \frac{\displaystyle (x)^{\frac{1}{2}} \frac{d}{dx}(y)^{\frac{1}{2}} - (y)^{\frac{1}{2}} \frac{d}{dx} (x)^{\frac{1}{2}} }{[(x)^{\frac{1}{2}}]^2} \\ \\ \\ \\ \frac{d^2 y}{dx^2} =& - \frac{\displaystyle (x)^{\frac{...
Read more

What was the effect of World War II on African Americans?

According to historians John Hope Franklin and Alfred A. Moss, Jr., more than three million black men registered for the Selective Service. However, they were refused at a rate of 18.2 percent, compared to just 8.5 percent for whites.  Discrimination was rampant in the armed forces, though black soldiers had more opportunities than in the First World War. They were included in the infantry, the coast and field artillery, the engineer corps, the medical corps, and many other branches. When the Women's Army Corps was organized (WAC), black women were included. By the end of the war, more than 4,000 women had enlisted in the organization. The next step was to get black soldiers recommended for advanced training at officer training schools. Blacks were not admitted to officer training schools in high numbers until the Secretary of War issued an order stating that blacks were to be admitted. The integration of the military began in 1945, first on the war front in Germany. In 1948, sever...
Read more