# Thread: equation of tangent line

1. ## equation of tangent line

Find the equation of the tangent line at the given value (in form Ax+By+C=0) Implicit Differentiation must be used:

If you could show me the steps, as well, that would be great... yet again, i'm totally lost.

a) x^2 y^3 = 8 ; (-1,2)

b) y^3 + xy - y = 8x^4 ; x = 1

thanks again, guys :-/

2. Hello, shepherdm1270!

Here's the first one . . .

Find the equation of the tangent line at the given value
(in the form $Ax+By+C\:=\:0$).
Implicit Differentiation must be used:

$a)\;\;x^2y^3 \:=\:8\qquad(\text{-}1,2)$

Differentiate: . $x^2\!\cdot\!3y^2y' + 2x\!\cdot\!y^3 \:=\:0\quad\Rightarrow\quad y' \:=\:-\frac{2y}{3x}$

At $(\text{-}1,2)\!:\;\;y' \:=\:-\frac{2(2)}{3(\text{-}1)} \:=\:\frac{4}{3}\quad\hdots$ The slope of the tangent is $\frac{4}{3}$

The line through $(\text{-}1,2)$ with slope $\frac{4}{3}$ is:

. . $y - 2 \;=\;\frac{4}{3}[x - (\text{-}1)]\quad\Rightarrow\quad y - 2 \:=\:\frac{4}{3}x + \frac{4}{3}$

Multiply through by $3\!:\;\;3y - 6 \:=\:4x + 4\quad\Rightarrow\quad \boxed{ 4x - 3y + 10 \:=\:0}$

3. Originally Posted by shepherdm1270
b) $y^3 + xy - y = 8x^4$ ; x = 1
First of all if you sub $x = 1$ you'll get
$y^3=8$, which means $y=2$

and if we differentiate:
$y^3 + xy - y = 8x^4$

$3y^2\frac{dy}{dx} + x\frac{dy}{dx} + y - \frac{dy}{dx} = 32x^3$

$\frac{dy}{dx}(3y^2 + x - 1) = 32x^3 - y$

$\frac{dy}{dx} = \frac{32x^3 - y}{3y^2 + x -1}$

sub $x=1$ and $y=2$ you'll get $\frac{dy}{dx} = \frac{5}{2}$

$y-2 =\frac{5}{2}(x-1)$

$2y-4=5x-5$

$5x-2y-1=0$