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

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

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2. Originally Posted by Bilbo94

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$f(x) = 3x^2 - 4xy + 3y^3 = -1.3 e + 3$
$f'(x) = 6x - 4y - 4x \frac{dy}{dx} + 9 y^2 \frac{dy}{dy} = -1.3e$
$\implies 6x - 4y +1.3e = 4x \frac{dy}{dx} - 9y^2 \frac{dy}{dy}$
$6x - 4y +1.3e = \frac{dy}{dx} (4x - 9y^2)$
$\frac{dy}{dx} = \frac{6x - 4y + 1.3e}{4x - 9y^2}$

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

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$2x^2 - 4xy +3y^3 = -1.3e + 03$
$2x^2 - 4sy + 3y^3 = -1.3 \times 10^3$?