Donna Flynn

2022-07-23

I'm stuck on this problem involving implicit differentiation.

The instructions ask me to find ${y}^{\prime}$, the problem is:

$(x-2y{)}^{3}=2{y}^{2}-3$

So far I've been able to get this far:

$3(x-2y{)}^{2}(1-2{y}^{\prime})=4y({y}^{\prime})$

I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.

The instructions ask me to find ${y}^{\prime}$, the problem is:

$(x-2y{)}^{3}=2{y}^{2}-3$

So far I've been able to get this far:

$3(x-2y{)}^{2}(1-2{y}^{\prime})=4y({y}^{\prime})$

I've been trying to manipulate it for a while but I can't figure out how to finish the problem properly.

Monica Dennis

Beginner2022-07-24Added 13 answers

Another way could be to consider the implicit function

$f=(x-2y{)}^{3}-2{y}^{2}+3=0$

Compute its derivative with respect to $x$ and $y$

${f}_{x}^{\prime}=3(x-2y{)}^{2}$

${f}_{y}^{\prime}=-6(x-2y{)}^{2}-4y$

Now, use the implicit function theorem

$\frac{dy}{dx}=-\frac{{f}_{x}^{\prime}}{{f}_{y}^{\prime}}=-\frac{3(x-2y{)}^{2}}{-6(x-2y{)}^{2}-4y}=\frac{3}{2}\times \frac{(x-2y{)}^{2}}{3(x-2y{)}^{2}+2y}$

$f=(x-2y{)}^{3}-2{y}^{2}+3=0$

Compute its derivative with respect to $x$ and $y$

${f}_{x}^{\prime}=3(x-2y{)}^{2}$

${f}_{y}^{\prime}=-6(x-2y{)}^{2}-4y$

Now, use the implicit function theorem

$\frac{dy}{dx}=-\frac{{f}_{x}^{\prime}}{{f}_{y}^{\prime}}=-\frac{3(x-2y{)}^{2}}{-6(x-2y{)}^{2}-4y}=\frac{3}{2}\times \frac{(x-2y{)}^{2}}{3(x-2y{)}^{2}+2y}$

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