Question

If $f\left(x\right)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\infty$, then show that $f^{\prime }\left(x\right)=\sinh x$.

Answer 1

$f\left(x\right)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\infty$

Differentiate it

$f^{\prime }\left(x\right)=0+\frac{{2x}^{}}{2!}+\frac{4x^3}{4!}+\frac{6x^5}{6!}+...\infty$

$=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...+\infty =\sinh x$

Remember the expansion: $\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...+\infty$