The question is:

Find the equation of the tangent to the curve $y^2=8x$ at a point $(x_0,y_0)$.

My teacher's approach :

Differentiate the equation. we get $2y\cdot \dfrac {dy}{dx}=8 $

which gives us : $\dfrac {dy}{dx}=\dfrac4y$.

At a point $(x_0,y_0)$, slope of the tangent =$\dfrac4{y_0}$

So the equation of the tangent is $\dfrac{y-y_0}{x-x_0}=\dfrac4{y_0}$

What bothers me here is the differentiation. According to wikipedia :

the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point.

But, $y^2=8x$ is NOT a function. Then, how can one find the derivative? This bothers me a lot.. Can someone help me? Thank you.

  • $\begingroup$ It is a function, but $x=f(y)$. $\endgroup$ – Sigur Jul 25 '13 at 15:06
  • $\begingroup$ then, shouldnt we find $dx/dy$ instead of $dy/dx$? $\endgroup$ – user76849 Jul 25 '13 at 15:11
  • $\begingroup$ Good question! But don't forget that the slope of the tangent line is computed as the angle between it and the $x$-axis. So if you take $x'=y/4$ you have to invert it. $\endgroup$ – Sigur Jul 25 '13 at 15:13
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    $\begingroup$ Take for example the point $(x_0,y_0)=(9,-3)$. There is a unique function $y(x)$, defined for a (generous) interval about $x=9$, such that $y^2=x$. It is is the function usually called $-\sqrt{x}$. $\endgroup$ – André Nicolas Jul 25 '13 at 15:21
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    $\begingroup$ @user76849 For this particular problem, Andre Nicolas already told you the proof... you can solve for $y=f_+(x) = \sqrt{8x}$ or $y=f_{-}=-\sqrt{8x}$ depending on the point in question. The place where these glue together, $(0,0)$ is the one place this calculation fails. At $(0,0)$ the tangent to the curve is vertical and consequently falls outside the scope of functions of $x$. For the proof, there are many places, just about every advanced calculus text contains a proof. $\endgroup$ – James S. Cook Jul 25 '13 at 17:16

Forget about $y$ and start writing $f(x)^2=8x$. Now you do have a function ( it's just a matter of which symbols you are using ).

Derive this function in respect to $x$, in the left side you have $2f '(x)f(x)$, using the chain rule, and in the right side you have $8$. Both sides are equal, so you get $2f '(x)f(x)=8$, therefore $f '(x)=4/f(x)$.

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  • $\begingroup$ well, according to me $f(x)^2=8x$ is NOT a function. that's what the whole issue was. your answer doesnt answer my query. $\endgroup$ – user76849 Jul 25 '13 at 15:37
  • $\begingroup$ I'm sorry but I disagree, it IS a function. $f(x)$ is the function, coulbe you be more specific about why is not function according to you? $\endgroup$ – diff_math Jul 25 '13 at 15:38
  • $\begingroup$ @user76849 The left hand side of the equation is the rule of a function of $x$. So is the right. Since the two functions are equal, so are their derivatives. $\endgroup$ – David Mitra Jul 25 '13 at 15:42

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