I'm doing an AP Math Course and I ran into this problem that I have never seen before and would like some help and an explanation...
Consider the implicit function $\cosh(xy) = x + y$. Find $\frac{dy}{dx}$
At first it seemed simple enough, but the answer choices that are given are what confuses $$ \newcommand{\csch}{\text{csch}} \begin{eqnarray} A\qquad &\frac{xy-\csch(xy)}{y-\csch(xy)}\\ B\qquad &\frac{\csch(xy) - xy}{1-\csch(xy)}\\ C\qquad &\frac{\csch(xy)-y}{x-\csch(xy)}\\ D\qquad &\frac{\csch(xy)-y}{\csch(xy) + x} \end{eqnarray} $$
I've never seen $h$ before so what does that mean? I feel like I missed a chapter in the textbook but just in case, can someone explain that and also how to find the answer?
The functions $\cosh$ and $\sinh$ are known as hyperbolic functions. The definitions are: $$\cosh x = \frac{e^x + e^{-x}}{2} \qquad \quad \sinh x = \frac{e^x - e^{-x}}{2} $$ It is easy to remember the signs, thinking that $\cos$ is an even function, and $\sin$ is odd. You can prove easily using the definitions above that $\sinh' = \cosh$ and $\cosh' = \sinh $ (no minus sign here. We define $\tanh, \mathrm{sech}$, etc by the obvious way. Whenever doing implicit differentiation, it is nice to write clearly who is function of who. Writing $y = f(x)$, we have: $$\begin{align} \cosh(x \cdot f(x)) &= x + f(x) \\ (x \cdot f'(x) + f(x)) \sinh(x \cdot f(x)) &= 1 + f'(x) \\ x \cdot f'(x) \sinh(x \cdot f(x)) - f'(x) &= 1 - f(x) \cdot \sinh(x \cdot f(x)) \\ f'(x) = \frac{1 - f(x)}{x \sinh(x \cdot f(x)) - 1} \end{align}$$
Now it's just a matter of leaving the answer in a way we can choose an alternative. By definition, we can check that $$\cosh^2 t - \sinh^2 t = 1$$ Dividing by $\cosh^2 t $ and $\sinh^2 t $ we obtain $$\begin{align} 1 - \tanh^2 t &= \mathrm{sech}^2 t \\ \mathrm{coth}^2 t - 1 &= \mathrm{csch}^2 t \end{align}$$ Going back to our answer and using these identities, you should manage to find the correct alternative, ok? (I would try to divide both the numerator and the denominator by something, now)