# Math Help - Slope of the tangent line to the curve

1. ## Slope of the tangent line to the curve

at

Please help. I'm so confused

2. Originally Posted by Bilbo94

at

Please help. I'm so confused
$f(x) = 3x^2 - 4xy + 3y^3 = -1.3 e + 3$
$f'(x) = 6x - 4y - 4x \frac{dy}{dx} + 9 y^2 \frac{dy}{dy} = -1.3e$
$\implies 6x - 4y +1.3e = 4x \frac{dy}{dx} - 9y^2 \frac{dy}{dy}$
$6x - 4y +1.3e = \frac{dy}{dx} (4x - 9y^2)$
$\frac{dy}{dx} = \frac{6x - 4y + 1.3e}{4x - 9y^2}$

Gradient found when $(5 , -8)$ substituted into $\frac{dy}{dx}$.

$Gradient = -0.118$ to $3sf$

Equation of tangent:
$y - (-8) = -0.118 (x - 5)$
$y = -0.118x -7.41$

3. I followed all your steps, but the homework module still says it's wrong.

4. Originally Posted by Bilbo94

at

Please help. I'm so confused
Even assuming that Air's calculations are correct, there is still something funny about how this is written.
$2x^2 - 4xy +3y^3 = -1.3e + 03$

Could this possibly be
$2x^2 - 4sy + 3y^3 = -1.3 \times 10^3$?

-Dan