How would you differentiate this implicitly twice?
dy/dx = e^xy

Thanks.

You could try taking (natural) logs first. When you differentiate ln(dy/dx), you can differentiate with respect to (dy/dx), and then multiply by (dy/dx differentiated with respect to x), which is d² y/dx² .

Hi Emma,

Can you show me what you describe?
Thanks again.

How do you differentiate ln(dy/dx) with respect to (dy/dx)?

Do you know how to differentiate log t with respect to t? That's 1/t.
The chain rule says that if you want to differentiate f(t) with respect to x, you do (df/dt) x (dt/dx), so that's (1/t) x (dt/dx).
In this case, t=dy/dx, so we get (1/(dy/dx)) x (d² y/dx² ).

Do come back to us if you get stuck further on. Multiplying through by dy/dx will be helpful, and then you'll have to keep very clear about the product rule.

Here is the complete working:

${\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=e^{xy}}$

${\displaystyle \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=xy}$

${\displaystyle \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}\frac{d^{2}y}{{\mathrm{dx}}^2}=y+x\frac{\mathrm{dy}}{\mathrm{dx}}}$

${\displaystyle \frac{d^{2}y}{{\mathrm{dx}}^2}=y\frac{\mathrm{dy}}{\mathrm{dx}}+x{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}$