How to find the implicit differentiation of a fraction?

Question

I need to determine the whether point P is a local max/min or stationary point. So I need to take the second derivative.

Question: $5x^2+6xy+5y^2 = 8$

I figured out that the first derivative is: $\frac{dy}{dx} = \frac{-10x-6y}{6x+10y}$

Therefore the second derivative is: $\frac{d^2y}{dx^2}$ = $\frac{dy}{dx}( \frac{-10x-6y}{6x+10y})$

But I am not sure how to proceed. I know that you add $\frac{dy}{dx}$ every time you differetiate a $y$ (at least this is the way I have been taught, but how do you go about and do this? There are fractions and $x$ and $y$ variables are on top and bottom. Thanks

Answer 1

Instead of messing about with the quotient rule and all of the opportunities for making silly algebraic and arithmetic errors that presents, since you’re already differentiated the original implicit equation with respect to $x$ in the process of computing $y'$, just differentiate again: from $$10x+6y+6xy'+10yy' = 0$$ we get $$10+12y'+6xy''+10(y')^2+10yy'' = 0.$$ Now solve for $y''$ and substitute your already-computed expression for $y'$, or vice-versa.

Answer 2

It's fiddly and messy, but simple enough to use the quotient rule for derivatives:

$\mathrm d \left(\dfrac u v\right) = \dfrac {v \mathrm d u - u \mathrm d v} {v^2}$

You have, for example, $v = 6 x + 10 y$ which gives:

$\dfrac {\mathrm d v} {\mathrm d x} = 6 + 10 \dfrac {\mathrm d y} {\mathrm d x}$

and $u = -10 x - 6 y$, which gives:

$\dfrac {\mathrm d u} {\mathrm d x} = -10 - 6 \dfrac {\mathrm d y} {\mathrm d x}$

It remains to be assembled.

Answer 3

Differentiate, divide by 2 $$ 5 x + 3 ( xy' +y) + 5 yy' =0 \tag1 $$

$$ y'= \dfrac{-(5x+3y)}{3x+5 y} \tag2 $$

Differentiate (1) again

$$ 5 +3 (y'+ xy''+ y' ) +5(y y''+ y^{'2}) =0 \tag3 $$

Plug in from (2) for $y'$ and simplify to find y''.

Answer 4

The equation $ \ 5x^2 + 6xy + 5y^2 \ = \ 8 \ $ describes a "tilted" ellipse (see the graph at bottom). As such, then, this is not the equation of a function, but of two separate (implicit) functions (about which more shortly).

You appear to have correctly carried out the implicit differentiation $ \ 10x \ + \ 6y \ + \ 6x·\frac{dy}{dx} \ + \ 10y·\frac{dy}{dx} \ = \ 0 \ \ . $ You did not provide the coordinates of the point of interest on this curve, but we can locate the points where "horizontal tangents" occur by inserting $ \ \frac{dy}{dx} \ = \ 0 \ \ $ into this equation to obtain $ \ 10x \ + \ 6y \ = \ 0 \ \ ; $ the points on the ellipse concerned are the intersections with the line $ \ y \ = \ -\frac{10}{6}x \ : \ $ $$ 5x^2 \ + \ 6x·\left(-\frac53x \right) \ + \ 5·\left(-\frac53x \right)^2 \ \ = \ \ 8 \ \ \Rightarrow \ \ \frac{5·9 \ - \ 30·3 \ + \ 5·25}{9} \ · \ x^2 \ \ = \ \ 8 $$ $$ \Rightarrow \ \ \frac{80}{9} \ · \ x^2 \ \ = \ \ 8 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{72}{80} \ \ = \ \ \frac{9}{10} $$ $$ \Rightarrow \ \ x \ \ = \ \ \pm\frac{3}{\sqrt{10}} \ \ = \ \ \pm\frac{3\sqrt{10}}{10} \ \ , \ \ y \ \ = \ \ -\frac{10}{6}· \left(\pm\frac{3\sqrt{10}}{10} \right) \ \ = \ \ \mp\frac{\sqrt{10}}{2} \ \ . $$

Note that $ \ (10x \ + \ 6y) \ + \ (6x \ + \ 10y)·\frac{dy}{dx} \ = \ 0 \ \ $ can be regarded as a linear equation with $ \ \frac{dy}{dx} \ $ as the "unknown". So if you are given the coordinates of a point on the ellipse, you can insert them into the equation and solve for $ \ \frac{dy}{dx} \ $ directly. (You really only need to write the expression $ \ \frac{-10x-6y}{6x+10y} \ $ if a homework or exam problem specifically asks for it.) If you are given the slope of a tangent line and asked for the points at which that slope occurs, you can insert that value for $ \ \frac{dy}{dx} \ $ into the same equation to obtain the equation for the line which intersects the ellipse at those points.

[Warning: the process of implicit differentiation is a rather "open" device, in that it can be used to give derivatives at points not on the curve and derivative functions for non-existent curves (such as $ \ x^2 + y^2 \ = \ -4 \ \ ) \ , $ which are meaningless results.]

To find the points at which "vertical tangents" occur, we could use the relation that you found, $ \ \frac{dy}{dx} \ = \ \frac{-10x-6y}{6x+10y} \ \ , $ observing that it is "undefined" where the denominator is zero. So the points with vertical tangents are the intersections of the ellipse with the line $ \ 6x \ + \ 10y \ = \ 0 \ \ $ or $ \ y \ = \ -\frac{6}{10}x \ \ . $ Another way we could find this is by differentiating implicitly the curve equation with respect to $ \ \mathbf{y} \ \ , $ which produces $$ \ 10x· \frac{dx}{dy} \ + \ 6· \frac{dx}{dy}·y \ + \ 6x \ + \ 10y \ \ = \ \ 0 \ \ . $$ For a vertical tangent, $ \ \frac{dx}{dy} \ = \ 0 \ \ , $ so inserting this into our new equation yields $$ y \ = \ -\frac{6}{10}x \ \ \Rightarrow \ \ 5x^2 \ + \ 6x·\left(-\frac35x \right) \ + \ 5·\left(-\frac35x \right)^2 \ \ = \ \ 8 $$ $$ \Rightarrow \ \ \frac{5·5 \ - \ 6·3 \ + \ 3^2}{5} \ · \ x^2 \ \ = \ \ 8 \ \ \Rightarrow \ \ \frac{16}{5} \ · \ x^2 \ \ = \ \ 8 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{40}{16} \ \ = \ \ \frac{10}{4} $$ $$ \Rightarrow \ \ x \ \ = \ \ \pm\frac{\sqrt{10}}{2} \ \ , \ \ y \ \ = \ \ -\frac{6}{10}· \left(\pm\frac{\sqrt{10}}{2} \right) \ \ = \ \ \mp\frac{3\sqrt{10}}{10} \ \ . $$

[Note that the horizontal and vertical tangent points are symmetrically arranged about the line $ \ y \ = \ -x \ \ ; \ $ this is because the major axis of the ellipse lies on this line.

Also, as we should expect, the relations $ \ \frac{dy}{dx} \ = \ -\frac{10x+6y}{6x+10y} \ \ $ and $ \ \frac{dx}{dy} \ = \ -\frac{6x+10y}{10x+6y} \ \ $ are reciprocals of one another.

To return to an earlier point, the vertical tangent points "divide" the ellipse into two parts which are the implicit function curves of $ \ x \ \ . $ We can manage to obtain these by applying the quadratic formula to the ellipse equation:

$$ 5y^2 \ + \ 6xy \ + \ (5x^2 - 8) \ = \ 0 $$ $$ \Rightarrow \ \ y \ \ = \ \ \frac{-6x \ \pm \ \sqrt{36x^2 \ - \ 4·5·(5x^2 - 8)}}{2·5} \ \ = \ \ -\frac35 x \ \pm \ \frac{2\sqrt{10 \ - \ 4x^2 }}{5} \ \ . \ ] $$

Since you are asked to find the "concavity" of the ellipse at a point, we need the second derivative, which can be obtained by differentiating our first derivative equations implicitly once again (as shown by the other posters). Unless we require an expression for that, however, we can carry out the differentiation $$ 10·1 \ + \ 6·\frac{dy}{dx} \ + \ (6·1·\frac{dy}{dx} \ + \ 6x·\frac{d^2y}{dx^2}) \ + \ (10·\frac{dy}{dx}·\frac{dy}{dx} \ + \ 10·y·\frac{d^2y}{dx^2}) \ = \ 0 $$ $$ \Rightarrow \ \ 5 \ + \ 6·\frac{dy}{dx} \ + \ 5·\left(\frac{dy}{dx} \right)^2 \ + \ (3x \ + \ 5 y)·\frac{d^2y}{dx^2} \ = \ 0 $$

and then simply insert the coordinates and first derivative into this equation to calculate the second derivative (again, as a linear equation). For our horizontal tangent points, we have $ \ \frac{dy}{dx} \ = \ 0 \ \ , $ so we may compute $$ 5 \ + \ 6·0 \ + \ 5·0^2 \ + \ (3x \ + \ 5 y)·\frac{d^2y}{dx^2} \ \ = \ \ 0 \ \ \Rightarrow \ \ \frac{d^2y}{dx^2} \ \ = \ \ -\frac{5}{3x \ + \ 5y} \ \ , $$ giving us $$ \frac{d^2y}{dx^2} \ \ = \ \ -\frac{5}{3·\frac{3\sqrt{10}}{10} \ + \ 5·\left(-\frac{\sqrt{10}}{2} \right)} \ \ = \ \ \frac{25}{8\sqrt{10}} \ \ > \ \ 0 \ \ , $$ so the ellipse is "concave upward" [local and absolute minimum] at $ \ \left(\frac{3\sqrt{10}}{10} \ , \ -\frac{\sqrt{10}}{2} \right) \ \ , $ and $$ \frac{d^2y}{dx^2} \ \ = \ \ -\frac{5}{3·\left(-\frac{3\sqrt{10}}{10} \right) \ + \ 5· \frac{\sqrt{10}}{2} } \ \ = \ \ -\frac{25}{8\sqrt{10}} \ \ < \ \ 0 \ \ , $$ signifying "downward concavity" [local and absolute maximum] at $ \ \left(-\frac{3\sqrt{10}}{10} \ , \ \frac{\sqrt{10}}{2} \right) \ \ . $ (We see this confirmed in the graph.) We could perform similar calculations for the vertical tangent points using $ \ \frac{d^2x}{dy^2} \ \ . $