Implicit differentiation of $\log$ and $\sin$ function

Question

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

score 0 · Accepted Answer · answered Sep 02 '20 at 21:35

0

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?

answered Sep 02 '20 at 21:35

Ninad Munshi

I hope so, but it would be nice if you could provide a solution, so that I can check – Henrik Larsen Sep 02 '20 at 21:40
@HenrikLarsen post your stab at a solution first, then me or someone else can tell you if it's right or not – Ninad Munshi Sep 02 '20 at 21:41
I find $y''(0)=280$. Is that right? – Henrik Larsen Sep 02 '20 at 21:48
@HenrikLarsen yes that is correct – Ninad Munshi Sep 02 '20 at 21:56

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