Determine the value of g''(x) using implicit differentiation

Question

for part a I got that g(0) = 1 and for part b I got that g'(0) = 0. However, I am stuck on part c. I'm not sure if I did parts a or b incorrectly or if I'm on the right track.

Answer 1

Your answers for part a and b are correct.

For part c, taking the derivative of $(f(x))^2+(g(x))^2=1$ gives us $$2f(x)f'(x)+2g(x)g'(x)=0$$ which simplified is just $$f(x)f'(x)+g(x)g'(x)=0$$

Since the second derivative of both functions exist, we can take the second derivative giving us $$f'(x)f'(x)+f(x)f''(x)+g'(x)g'(x)+g(x)g''(x)=0$$

Simplifying this gives us $$(f'(x))^2+f(x)f''(x)+(g'(x))^2+g(x)g''(x)=0$$

Since $f'(x) = (g(x))^2$, we can substitute this in to get $$(g(x))^4+f(x)f''(x)+(g'(x))^2+g(x)g''(x)=0$$

Now plugging in $x=0$ and using part a and b, you can find what $g''(0)$ should be.