The correct choice is (a) 43
For explanation I would say: Expanding sinh(x) into a Taylor series we have
sinh(x)=\(\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty\)
Now rewriting the function we have
f(x)=(x^3+x^2+x+1)\((\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty)\)
For the 7^th derivative observe that, only the odd termed powers contribute to the derivative at x=0
Hence it is enough for us to find seventh derivative for
(x^2+1)\((\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty)\)
\(\frac{x}{1!}+x^3(\frac{1}{3!}+\frac{1}{1!})+x^5(\frac{1}{5!}+\frac{1}{3!})+…\infty\)
Taking the 7^th derivative of this function we have
f^(7)(x)=\((7!)(\frac{1}{7!}+\frac{1}{5!}) + (9*8…4*3)x^2(\frac{1}{7!}+\frac{1}{9!})\)
Now substituting x=0 yields
f^(7)(0)=\((7!)(\frac{1}{5!}+\frac{1}{7!})\)=(1+7*6)=43.