# Show that $y''+\tan y\cdot(y')^2=0$ when $y=\tan^{-1}(\sin h x)$

I have tried solving it in this manner: $$\tan y=\sin h x$$ Differentiate with respect to $x$: \begin{align*} \sec^2y_1&=h \cos hx\\ sec^2 y\frac{d y}{dx}&=h \cos hx\\ \frac{d y}{dx}&=\frac{h\cos hx}{\sec^2y}\\ y_1&=\frac{h\cos hx}{1+\tan^2y}\\ y_1+y_1(\tan^2y)=h\cos hx \end{align*} Differentiate with respect to $x$: \begin{align*} y_2+2y_1\tan y\sec^2y+y_2\tan^2y=-h^2\sin hx \\ y_2=\frac{-(h^2\sin hx+2y_1\tan y \sec^2y)}{1+\tan^2y}. \end{align*} This as far as I have gotten.

• Are you sure it's $\sin hx$ (sine of $hx$) and not $\sinh x$ (hyperbolic sine of $x$)? – egreg Sep 2 '14 at 22:09
• Are you sure $h$ is a constant? Because it looks like you're dealing with hyperbolic sine. – Mike Sep 2 '14 at 22:10
• In the question they have neither mentioned "constant" or "hyperbolic sine".i don't know what is hyperbolic sine, so i assumed "h" to be a constant.Please help me solve this. – Nikhil01 Sep 2 '14 at 22:13

If the equation is $y=\tan^{-1}(\sinh x)$, where $\sinh x$ is the hyperbolic sine of $x$, then we can compute $y'$ with the chain rule: $$y'=\frac{1}{1+\sinh^2 x}\cosh x=\frac{1}{\cosh^2x}\cosh x=\frac{1}{\cosh x},$$ recalling that $\cosh^2x-\sinh^2x=1$. Thus $$y''=-\frac{\sinh x}{\cosh^2x}$$ and the stated relation follows immediately.
The hyperbolic sine and cosine are defined by $$\sinh x=\frac{e^x-e^{-x}}{2},\qquad \cosh x=\frac{e^x+e^{-x}}{2}.$$ From the definition it's easy to compute $$\sinh'x=\cosh x,\quad \cosh'x=\sinh x,\quad \cosh^2x-\sinh^2x=1.$$
• how is $sinh'x=coshx$ and$cosh'x=sinhx$ and what does "$sinh'x$ or$sinh'x$ "signify?? – Nikhil01 Sep 2 '14 at 22:24
• @Nikhil01 With $\sinh'x$ I mean the derivative; since the derivatives of $e^x$ and $e^{-x}$ are $e^x$ and $-e^{-x}$ the stated formulas follow. The hyperbolic sine and cosine are quite useful when doing integrals. – egreg Sep 2 '14 at 22:27