# a line tangent to curve $y = \frac{x}{2−2x}$ goes through (1, -1)

$$y = \frac{x}{2−2x}$$

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

$$y = \frac{x}{2−2x}$$

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

why cant i find the gradient with derivative?

• 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. – irchans Oct 19 '19 at 8:57
• The derivative is the gradient for real valued functions of real numbers. – 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.

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 $$k=\frac18$$.

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