# Find the derivative of  x^2 + y^2 = (5x^2 + 4y^2 -x)^2 , and find the equation of the tangent at (0,0.25)?

Feb 24, 2017

The derivative is given (implicitly) by;

$x + y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

The equation of the tangent at $\left(0 , 0.25\right)$ is:

$y = x + \frac{1}{4}$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

${x}^{2} + {y}^{2} = {\left(5 {x}^{2} + 4 {y}^{2} - x\right)}^{2}$

Differentiate wrt $x$ (applying chain rule):

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

$\therefore x + y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$

Whilst we could spend time and effort to find an explicit equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we have no requirement to, so lets just find $\frac{\mathrm{dy}}{\mathrm{dx}}$ at the point $\left(0 , 0.25\right)$;

$\implies 0 + \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(0 + 4 \left(\frac{1}{16}\right) - 0\right) \left(0 + \frac{8}{4} \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{4}\right) \left(2 \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{4}$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{1}{4} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{4}$

$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

So the tangent passes through $\left(0 , 0.25\right)$ and has gradient $1$ so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek is;

$\setminus \setminus \setminus \setminus \setminus y - \frac{1}{4} = 1 \left(x - 0\right)$
$\therefore y - \frac{1}{4} = x$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus y = x + \frac{1}{4}$

We can verify this solution graphically;

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = {x}^{2} + {y}^{2} - {\left(5 {x}^{2} + 4 {y}^{2} - x\right)}^{2}$; Then;

$\frac{\partial F}{\partial x} = 2 x - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)$

$\frac{\partial F}{\partial y} = 2 y - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(8 y\right)$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)}{2 y - 2 \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(8 y\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{x - \left(5 {x}^{2} + 4 {y}^{2} - x\right) \left(10 x - 1\right)}{y - 8 y \left(5 {x}^{2} + 4 {y}^{2} - x\right)}$

Here we get an immediate implicit function for the derivative, so again at $\left(0 , 0.25\right)$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{0 - \left(0 + \frac{4}{16} - 0\right) \left(0 - 1\right)}{\frac{1}{4} - \frac{8}{4} \left(0 + \frac{4}{16} - 0\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{- \left(\frac{1}{4}\right) \left(- 1\right)}{\frac{1}{4} - 2 \left(\frac{1}{4}\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{\frac{1}{4}}{- \frac{1}{4}}$
$\setminus \setminus \setminus \setminus \setminus \setminus = 1 \setminus \setminus \setminus \setminus$, as before