1
$\begingroup$

Suppose we have a curve in the form $x=h(y) $ and for what ever reason we are unable to define $y=f(x)$. Also suppose we want to find the tangent line to the curve at the point $P (x_0,y_0)$.

Now, say I didn't want to use implicit differentiation. So, the next resort would be to find the derivative of x with respect to y. Doing this we get: $$ \displaystyle \frac{dx}{dy} = h'(y) $$

My understanding of the derivative is as follows, when we differentiate a function with respect to a variable, we are measuring the rate at which function changes with respect to that variable. In this example, we are finding $$ \displaystyle \lim_{c\to 0} \frac{h(y+c) - h(y)}{c} = h'(y) $$

Now, if I wanted to find the gradient of the tangent line at $P$, could I just calculate $h'(y_0)$ and then write the equation of the tangent line as $$y-y_0 = h'(y_0)(x-x_0)$$

I have a feeling I'm missing something really simple as to why the gradient of the tangent line can't be $h'(y_0)$

  • 1
    That would be the run over the rise rather than the rise over the run...the slope should be $1/h'(y)$. Alternatively use $x-x_0=h'(y_0)(y-y_0)$ and solve for $y=mx+c$. – 2017-02-28
  • 0
    Yep yep what Jp said. $\frac{dx}{dx}=h'(y)$... Flipping both sides gives: $\frac{dy}{dx}=\frac{1}{h'(y)}$. – 2017-02-28
  • 0
    Since $\displaystyle \frac{dx}{dy} = h'(y)$ then $\displaystyle \frac{dy}{dx} = \frac 1{h'(y)}$ may be. – 2017-02-28
  • 0
    Understood, thanks – 2017-02-28

0 Answers 0