# Math Help - Implicit Differentiation Slope Tangent Line

1. ## Implicit Differentiation Slope Tangent Line

use implicit differentiation to find the slope of the tangent line to the curve

y/(x-3y) = x^6 + 8 at the point (1, 9/28)

m = ?

i tried using quotient rule but cannot seem to get the correct answer. this is what i have done

[(dy/dx)(x-3y)-(1-3dy/dx)(y)]/(x-3y)^2 = 6x^5

--> [x(dy/dx)]/(x-3y)^2 = 6x^5 + y

dy/dx = [(6x^5 + y)(x-3y)^2]/x

2. $\frac{y}{x-3y} = x^6 + 8$

$y = x^7 + 8x - 3x^6y - 24y$

$25y = x^7 + 8x - 3x^6y$

$25\frac{dy}{dx} = 7x^6 + 8 - 3x^6 \frac{dy}{dx} - 18x^5y$

$\frac{dy}{dx} = \frac{7x^6 + 8 - 18x^5y}{3x^6 + 25}$

see if that works.

3. it did work thanks a lot