Answer
$\dfrac{dy}{dx}$ at $(0,\sqrt[4] 2) $ is equal to $\dfrac{-\sqrt[4] 2-1}{2^{\frac{11}{4}}}$.
$\dfrac{dy}{dx}$ at $(0,-\sqrt[4] 2) $ is equal to $\dfrac{\sqrt[4] 2-1}{-2^{\frac{11}{4}}}$.
The required equation of the tangent line is $y=\dfrac{x}{5}-\dfrac{6}{5}$.
Work Step by Step
Firstly find the two points by substituting $x=0$ in $y^4 + xy = x^3 − x + 2$.
we get, $y^4 + (0)y = 0^3 − 0 + 2$
Now simplify and find the y-coordinates of the two points.
$y^4 = 2$
$y=\pm\sqrt[4] 2$
So the points are $(0,\sqrt[4] 2)$ and $(0,-\sqrt[4] 2)$.
Now differentiate $y^4 + xy = x^3 − x + 2$ with respect to $x$ using product rule and chain rule.
We get, $4y^3\dfrac{dy}{dx} + y\dfrac{dx}{dx}+x\dfrac{dy}{dx} = 3x^2 − 1 $
Now take $\dfrac{dy}{dx}$ common.
We get, $(4y^3 +x)\dfrac{dy}{dx}+y = 3x^2 − 1 $
Now solve for $\dfrac{dy}{dx}$ as follow.
$(4y^3 +x)\dfrac{dy}{dx} = 3x^2 -y− 1 $
$\implies \dfrac{dy}{dx} = \dfrac{3x^2 -y− 1}{4y^3+x} $
To find $\dfrac{dy}{dx}$ at $(0,\sqrt[4] 2) $ substitute $x=0$ and $y=\sqrt[4] 2$ in $\dfrac{dy}{dx} = \dfrac{3x^2 -y− 1}{4y^3+x} $.
$\dfrac{dy}{dx} = \dfrac{3(0)^2 -\sqrt[4] 2− 1}{4(\sqrt[4]2)^3+0} =\dfrac{-\sqrt[4] 2-1}{4(\sqrt[4] 2)^3}=\dfrac{-\sqrt[4] 2-1}{2^{\frac{11}{4}}}$
To find $\dfrac{dy}{dx}$ at $(0,-\sqrt[4] 2) $ substitute $x=0$ and $y=-\sqrt[4] 2$ in $\dfrac{dy}{dx} = \dfrac{3x^2 -y− 1}{4y^3+x} $.
$\dfrac{dy}{dx} = \dfrac{3(0)^2 +\sqrt[4] 2− 1}{4(-\sqrt[4]2)^3+0} =\dfrac{\sqrt[4] 2-1}{4(-\sqrt[4] 2)^3}=\dfrac{\sqrt[4] 2-1}{-2^{\frac{11}{4}}}$
Now to find the equation of the tangent line at $(1,1)$, substitute $x=1$ and $y=1$ in $\dfrac{dy}{dx} = \dfrac{3x^2 -y− 1}{4y^3+x} $.
$\dfrac{dy}{dx} = \dfrac{3(1)^2 -1− 1}{4(1)^3+1} =\dfrac{1}{5}$.
Thus the slope of the tangent line at $(1,1)$ is $\dfrac{1}{5}$.
Using the slope-intercept form of the tangent line.
We get, $y-1=\dfrac{1}{5}(x-1)$
Now simplify to get the equation of the tangent line.
$y-1=\dfrac{x}{5}-\dfrac{1}{5}$
$\implies y=\dfrac{x}{5}-\dfrac{1}{5}-1$
$\implies y=\dfrac{x}{5}-\dfrac{6}{5}$
Hence, the required equation of the tangent line is $y=\dfrac{x}{5}-\dfrac{6}{5}$.
