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I am struggling with the following problem for implicit differentiation.

I am tasked to differentiate implicitly the following function, and evaluate $y''(0)$, where $y=y(x)$.

$$\ln(y+1)+\sin(xy)=\ln(5).$$

I have differentiated this once to find,

$$(y+xy')\cos(xy)+\frac{y'}{y+1}=0$$

But how to advance from here to find $y''(0)$?

Thanks

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In the original equation we can find $y(0)$:

$$\ln(y+1) + \sin 0 = \ln 5 \implies y(0) = 4$$

And in that equation we can find $y'(0)$:

$$(4+0)\cos 0 + \frac{y'}{5} = 0 \implies y'(0) = -20$$

Now just implicit differentiate that expression again and follow the same procedure. Can you take it from here?

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  • $\begingroup$ I hope so, but it would be nice if you could provide a solution, so that I can check $\endgroup$ – Henrik Larsen Sep 2 at 21:40
  • $\begingroup$ @HenrikLarsen post your stab at a solution first, then me or someone else can tell you if it's right or not $\endgroup$ – Ninad Munshi Sep 2 at 21:41
  • $\begingroup$ I find $y''(0)=280$. Is that right? $\endgroup$ – Henrik Larsen Sep 2 at 21:48
  • $\begingroup$ @HenrikLarsen yes that is correct $\endgroup$ – Ninad Munshi Sep 2 at 21:56

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