# How do you use implicit differentiation here? the answer should be (5x^4+4x^3y)/(6xy^2-4y^3) but I don't know how to get there

## ${x}^{4} \left(x + y\right) = {y}^{3} \left(3 x - y\right)$

Oct 30, 2016

That answer is not correct for this question. Whoever provided the answer made an error in the product rule. (Or they started with a different equation.)

#### Explanation:

The given equation is equivalent to

${x}^{5} + {x}^{4} y = 3 x {y}^{3} - {y}^{4}$

Differentiating with respect to $x$ we will get 6 terms. Three involving $\frac{\mathrm{dy}}{\mathrm{dx}}$ and three not involving the derivative of $y$. Unless some of them combine, $\frac{\mathrm{dy}}{\mathrm{dx}}$ will be the ratio of two trinomials not two binomials.

$5 {x}^{4} + 4 {x}^{3} y + {\underbrace{{x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}}}}_{\text{this term is missing") = underbrace(3y^3)_("this term is missing") +underbrace(3x*3y^2 dy/dx)_("this term is incorrect}} - 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

$5 {x}^{4} + 4 {x}^{3} y + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {y}^{3} + 9 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

$5 {x}^{4} + 4 {x}^{3} y - 3 {y}^{3} = 9 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}}$

Oct 30, 2016

The correct answer (verified) is as follows:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4} + 4 {x}^{3} y - 3 {y}^{3}}{9 x {y}^{2} - 4 {y}^{3} - {x}^{4}}$

#### Explanation:

Implicit differentiation is just an application of the chain rule. If $y$ is a function of $x$ ( ie $y = f \left(x\right)$) then your differentiate to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ directly. But if y is not an explicit function of $x$ (eg ${y}^{2} = f \left(x\right)$), then you cannot find $\frac{\mathrm{dy}}{\mathrm{dx}}$ directly and instead differemtiate ${y}^{2}$ with wrt $y$ using the chain rule

So we have ${x}^{4} \left(x + y\right) = {y}^{3} \left(3 x - y\right)$

I would tackle this by first multiplying out:
${x}^{5} + {x}^{4} y = 3 x {y}^{3} - {y}^{4}$

We then differentiate everything wrt $x$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{5}\right) + \frac{d}{\mathrm{dx}} \left({x}^{4} y\right) = \frac{d}{\mathrm{dx}} \left(3 x {y}^{3}\right) - \frac{d}{\mathrm{dx}} \left({y}^{4}\right)$
$\therefore 5 {x}^{4} + \frac{d}{\mathrm{dx}} \left({x}^{4} y\right) = 3 \frac{d}{\mathrm{dx}} \left(x {y}^{3}\right) - \frac{d}{\mathrm{dx}} \left({y}^{4}\right)$

We will now also need to apply the product rule
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$ to get

$5 {x}^{4} + \left\{\left({x}^{4}\right) \left(\frac{d}{\mathrm{dx}} y\right) + \left(y\right) \left(\frac{d}{\mathrm{dx}} {x}^{4}\right)\right\} = 3 \left\{\left(x\right) \left(\frac{d}{\mathrm{dx}} {y}^{3}\right) + \left({y}^{3}\right) \left(\frac{d}{\mathrm{dx}} x\right)\right\} - \frac{d}{\mathrm{dx}} \left({y}^{4}\right)$

$5 {x}^{4} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{3} y = 3 \left\{x \left(\frac{d}{\mathrm{dx}} {y}^{3}\right) + {y}^{3}\right\} - \frac{d}{\mathrm{dx}} \left({y}^{4}\right)$

Now we can't differentiate functions of y wrt x but we can use the chain rule (and this is the implicit differentiation)

$5 {x}^{4} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{3} y = 3 \left\{x \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{3}\right) + {y}^{3}\right\} - \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{4}\right)$

This subtle change allows s now to differentiate the non-explicit functions of y;

$5 {x}^{4} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{3} y = 3 \left\{x \frac{\mathrm{dy}}{\mathrm{dx}} \left(3 {y}^{2}\right) + {y}^{3}\right\} - \frac{\mathrm{dy}}{\mathrm{dx}} \left(4 {y}^{3}\right)$

$\therefore 5 {x}^{4} + {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{3} y = 9 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{3} - 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

we can gather up the $\frac{\mathrm{dy}}{\mathrm{dx}}$ term on the LHS to get;
$5 {x}^{4} + 4 {x}^{3} y - 3 {y}^{3} = 9 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{4} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \left(9 x {y}^{2} - 4 {y}^{3} - {x}^{4}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} + 4 {x}^{3} y - 3 {y}^{3}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{4} + 4 {x}^{3} y - 3 {y}^{3}}{9 x {y}^{2} - 4 {y}^{3} - {x}^{4}}$