$\sin(y(x))+6x(y(x)) = 0$ and $y(0) = \pi$; what is $ y'(0)$?

Question

I need to find what $y'(0)$ is. What I know is that $\sin(y(x))+6x(y(x)) = 0$ and that $y(0) = \pi.$

When I derived it, I got $y'=\dfrac {\sin(y)}{x\cos(y)-6x^2}$ which would indicate that $y'(0) = 0.$ But that's apparently not correct. I've searched a lot and don't know how to solve this problem. Would be grateful if somebody could tell me how to solve it.

Answer 1

I believe that you may have misused the chain rule. Remember that $y$ is a function of $x$ so must be treated accordingly. Also note that at $x=0$ your denominator is $0$, which means that your $y'(0)$ is actually undefined.

Using the chain rule and product rule, and implicitly differentiating we get that: $$\cos(y(x))y'(x)+6y(x)+6xy'(x)=0.$$ It's important to note that, because $\sin(y(x))$ is a function composed with a different function of $x$ we have to use the chain rule here, and because $6xy(x)$ is the product of a function of $x$ (in this case $6x$), with a different function of $x$ (in this case $y(x))$ we must use the product rule here. Now if we substitute $x=0$ with the fact that $y(0)=\pi$ we find that $$-y'(0)+6\pi=0.$$

Answer 2

Differentiate this implicit function with respect to $x$, on both sides to get: $$\cos y(x)\frac{dy(x)}{dx}+6x\frac{dy(x)}{dx}+6y(x)=0$$

$$\frac{dy(x)}{dx}=\frac{-6y(x)}{6x+\cos y(x)}$$ Put $x=0$ to get, $$y'(0)=\frac{-6y(0)}{6(0)+\cos y(0)}=\frac{-6\pi}{-1}=6\pi$$

Hope this helps...