calculusderivativesimplicit-differentiation
Does the tangent line to the curve $x^3+xy^2+x^3y^5=3$ at the point $(1,1)$ pass through the point $(-2,3)$? (using implicit differentiation)
I got the implicit differentiation as $\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$. Then I got stuck?
Explanation would be appreciated.
If you want a horizontal tangent, then you need to look for where ${dy\over dx}=0$.
Edit: As indicated, solve ${dy\over dx}=0\implies y(2.5-1/x)=0\implies y=0\text{ or }x={2\over 5}$. But $y=0$ is not in the domain of the original relation, $\ln(xy)=2.5x$, so the horizontal tangent occurs only when $x=2/5\implies y=5e/2$.
This is depicted graphically below: the dashed line is the horizontal tangent passing thru $(2/5,5e/2)$ on the graph of $\ln(xy)=2.5x$.
[1] $x^2 + xy + y^2 = 1$
differentiating implicitly by $\frac{d}{dx}$
$2x + x\frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$
Which simplifies to
[2] $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$
Now consider a tangent linear equation which passes through a point on the curve and through $(1,1)$
It has the form
[3] $y = mx + c$
To find the points that these equations [1] and [3] intersect we must solve them simultaneously.
Substituting $y$ into [1]
$x^2 + x(mx+c) + (mx+c)^2 = 1$
This simplifies to
$(m^2+m+1)x^2 + (2mc+c)x + c^2 = 1$
using the quadratic formula
[4] $x = \frac{-2mc-c \pm \sqrt{(2mc+c)^2 - 4(m^2+m+1)c^2}}{2(m^2+m+1)}$
Note that there are 2 solutions because there are two points on the curve where this will work.
Also, since we know that equation [3] passes through the point $(1,1)$
we know $1= m1+c$
So [5] $m = 1 - c$
Se expression [4] becomes
[6] $x = \frac{-2(1 - c)c-c \pm \sqrt{(2(1 - c)c+c)^2 - 4((1 - c)^2+(1 - c)+1)c^2}}{2((1 - c)^2+(1 - c)+1)}$
Lets call these $x_1$ and $x_2$ They have a corresponding $y_1$ and $y_2$
We also know that the gradient of [3] is $m$. This will be equal to the derivative given by [2] at $(x_1, y_1)$ and $(x_2,y_2)$
ie:
[7] $m = \frac{-2x_1 - y_1}{x_1 + 2y_1} = 1 - c$
and
[8] $m = \frac{-2x_2 - y_2}{x_2 + 2y_2} = 1 - c$
Substitute [6] into [7] to get an expression for $y_1$ in terms of c Then you can substitute $x_1$ and $y_1$ into [1] to find what $c$ is. From here you can calculate $m$ which will give you the tangent line you are looking for.
A similar process can be used for $x_2$ and $y_2$ there are two tangent lines which will solve this problem.
I'll leave the rest for you
Best Answer
You've got one too many $-y^2$ terms in your numerator: you should arrive at $$\frac{dy}{dx}=\frac{-3x^2-y^2-3x^2y^5}{2xy+5x^3y^4}$$
With the correct derivative, evaluate $\dfrac {dy}{dx}$ at the point $(1, 1)$. That will give you the slope $m$ of the line tangent to your curve at $(1, 1)$. Then using $m$ and the point $(x_0, y_0) = (1, 1)$, you can write the equation of your line in point-slope form:
$$y - y_0 = m(x - x_0)\tag{1}$$
Once you've obtained your equation, then plug in the coordinates of $(-2, 3)$ into your equation of the tangent line $(1)$ to see if the equation, given $x = -2, y = 3$ is true. If so, then $(-2, 3)$ is on the line $(1)$ tangent to your curve at $(1, 1)$. If not, then it is not on that line $(1)$.