Differentiating $y = x - \frac2x + \frac3{x^2}$

Question

Another easy question for you guys.

I'm differentiating the below to find the equation of the tangent at $(-3,-2)$ $$y = x - \dfrac{2}{x} + \dfrac{3}{x^2}$$

I simplified to:

$$ y = x - 2x^{-1} + 3x^{-2}$$

Then differentiated to get:

$$ \frac{dy}{dx} = 1 + 2x^{-2} - 6x^{-3}$$ or $$\frac{dy}{dx} = 1 + \frac{2}{x^2} - \frac{6}{x^3}$$

Placing $x = -3$ into this gives me $1$, and placing $m=1$ into $y=mx+c$ gives me $c = 1$.

Making the simple equation: $0 = x - y + 1$

However, I'm given the answer as: $0 = 13x - 9y + 21$

Where did I go wrong, I've studied it for longer than I'm willing to admit, have I made a stupid mistake somewhere?

Thanks,

Answer 1

Your calculation up to $$\frac{dy}{dx} = 1 + \frac{2}{x^2} - \frac{6}{x^3}$$ is correct. However, placing $x=-3$ we get $$m=1+\frac{2}{(-3)^2}-\frac{6}{(-3)^3}=\frac{13}{9}$$ and so the tangent's equation is given by $$y - (-2) = \frac{13}{9}(x-(-3))$$ i.e. $$y = \frac{13}{9}x + \frac{7}{3}$$ or equivalently, $$9y - 13x - 21 = 0.$$