Using Newton's Method I am taking a Single Variable Calculus class and am stuck on a homework question. This is a two part question. I'm stuck on part B.
Part A) If the tangent line of $y =\cosh(x)$ at $x = a$ goes through the origin, what equation must $a$ satisfy?
Ans: $dy/dx = \sinh(x)$
A tangent line through the origin has   equation $y = mx$. If it meets the graph at $x = a$, then $ma = \cosh(a)$ and $m = \sinh(a)$. Therefore, $a\sinh(a) = \cosh(a)$.
Part B) Use Newton's Method to solve for $a$
 A: You are looking for the zero of function
$$f(a)=a \sinh(a)-\cosh(a)$$ for which
$$f'(a)=a \cosh (a)\qquad \text{and} \qquad f''(a)=a \sinh (a)+\cosh (a)$$ Since the function is even, you have two symmetric solutions.
What you can notice is that $f(0)=-1$, $f'(0)=0$ and $f''(0)=1$. So, to generate a guess, perform a Taylor expansion around $a=0$; this would give
$$f(a)=-1+\frac{a^2}{2}+\frac{a^4}{8}+O\left(a^6\right)$$ which is a quadratic in $a^2$. So, an estimate is 
$$a_0=\sqrt{2 \left(\sqrt{3}-1\right)}$$ Now, use Newton method which generates as itegrates
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}=a_n+\frac{1}{a_n}-\tanh (a_n)$$ which should converge very fast.
A: First you have to know something about how the function behaves by using a plotting programm, because Newton's method doesn't work for every function with any starting value.
$$a\cdot sinh(a)=cosh(a) \Rightarrow a \cdot sinh(a)-cos(a)=0$$
$$f(x)=x \cdot sinh(x)-cos(x)$$
If you plot the function you can see the function equals zero approx. in two points: -1 and 1 .
Suitable starting value for Newton's method would be 2 and -2.
Newton's method says $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
Thus for the function f:
$$x_{n+1}=x_n-\frac{x_n \cdot sinh(x_n)-cos(x_n)}{x_n\cdot cosh(x_n)}$$
$$x_0=2 \Rightarrow x_1 = 1.53597$$
$$x_1=1.53597 \Rightarrow x_2=1.27558$$
$$x_2=1.27558 \Rightarrow x_3=1.20423$$
$$x_3=1.20423 \Rightarrow x_4=1.1997$$
$$x_4=1.1997 \Rightarrow x_5=1.19968$$
and so on, after few iterations you will get estimate value accurate enough.
$$ x=1.19968 \Rightarrow f(x) = 0 $$
This is one solution, you can use same way to solve the other with starting value -2.
Full solution is then $$a= \pm 1.19968$$ .
