$y = \frac{x}{2−2x}$

$y = \frac{f(x)}{g(x)} \rightarrow y' = \frac{f'(x) \cdot g(x) - g'(x)\cdot f(x)}{{g(x)}^2}$

$y = \frac{x}{2−2x}$

$y' = \frac{(2-2x)-(x \cdot (-2))}{{(2−2x)}^2}$ $\rightarrow$ $y' = \frac{(2-2x+2x)}{{(2−2x)}^2} = \frac{2}{{(2-2x)}^2}$

why cant i find the gradient with derivative?

  • $\begingroup$ You have computed the derivative, so that means you know what the local slope of the curve is at (1,-1). If you have a point and a slope, you can figure out the equation of the line. $\endgroup$ – irchans Oct 19 '19 at 8:57
  • $\begingroup$ The derivative is the gradient for real valued functions of real numbers. $\endgroup$ – John Douma Oct 19 '19 at 8:58

Note that the point $(\color{red}1,\color{blue}{-1})$ does not belong to the curve $y=\frac{x}{2-2x}$. (Maybe, that is what puzzled you.) The tangent line to the curve must pass through this external point.

Let $\left(\color{red}{x_0},\color{blue}{\frac{x_0}{2-2x_0}}\right)$ be the tangent point on the curve. You must find the gradient (slope) of the tangent line at the point $\color{red}{x_0}$: $$f'(x_0)=\frac{2}{(2-2x_0)^2}=\frac{1}{2(1-x_0)^2}$$ Hence: $$\frac{\color{blue}{\frac{x_0}{2-2x_0}}-(\color{blue}{-1})}{\color{red}{x_0}-\color{red}1}=\frac2{(2-2x_0)^2} \Rightarrow x_0=3 \Rightarrow f'(x_0)=\frac18$$ So, the tangent line is: $y=\frac18x-\frac98$. See Desmos graph.


Your derivative is absolutely correct.

Equation of a line is $(y-y_1)=m(x-x_1)$ where $(x_1,y_1) $ is a point that lies on the line

Here $m = \frac{2}{(2-2x)^2}$

Now we know that $(1,-1)$ lies on the line and the gradient of the line is $\frac{2}{(2-2k)^2}$

Substitute everything

We get $y=\frac{2}{(2-2k)^2}(x-1)-1$

Now this line also passes from point and $(k,\frac{k}{2-2k})$

We get $\frac{k}{2-2k}=\frac{2}{(2-2k)^2}(k-1)-1$

Now solve this equation for k and you'll get $k=3$


The line through $(1,-1)$ is

  • $y=k(x-1)-1 \implies y'=k$

then we need to solve the two equations

  • $f(x_0)=\frac{x_0}{2−2x_0}=k(x_0-1)-1$
  • $f'(x_0)=\frac{1}{2(1-x_0)^2}=k$

The solution should be $x_0=3$ and $k=\frac18$.

  • $\begingroup$ Maybe, downvoter (not me) saw $y'=kx$? $\endgroup$ – farruhota Oct 19 '19 at 10:49
  • $\begingroup$ Ah ok! Thanks I can’t find that! Bye $\endgroup$ – user Oct 19 '19 at 10:51

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