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I was looking at the Newton's method at Wikipedia, and it is said that the equation of the tangent line to the curve $y = f(x)$ at $x = x_n$ is

$$y = f'(x_n) (x - x_n) + f(x_n)$$

The usual equation of a line is

$$y = m x + b$$

where $m$ is the slope, that is $\frac{\Delta y}{\Delta x}$, and $b$ is the "height" from where the line starts. I understand that $f'(x_n)$ is the slope of the tangent line (by definition), but why is it multiplied by $(x - x_n)$?

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  • $\begingroup$ If you're on the point $(x_n,y_n)$ on a line, and you change the $x$ coordinate by $\Delta x= x-x_n$, then the $y$ coordinate changes by $\Delta y=m\Delta x$. So the new $y$ coordinate is $y=m \Delta_x+ y_n$. This is your equation. $\endgroup$ – David Mitra Apr 9 '16 at 10:18
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Well as $m=f'(x_n)$ you get an equation $y = f'(x_n)x + b$.

Then you need to choose $b$ such that the line goes through $(x_n, f(x_n))$. If you choose $b = $f(x_n) - f'(x_n) x_n$ you achieve exactly that, and then we get:

$$y=mx+q = f'(x_n) x +f(x_n)-f'(x_n)x_n = f'(x_n)(x-x_n) + f(x_n)$$

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The equation of a straight line is given by $$y=ax+b$$ for some real $a$ and $b$. Now, the equation of the tangent to the graph of $f$ at the point $(x_n,f(x_n))$ is the straight line with $a=f'(x_n)$, so we get $$y=f'(x_n)+b.$$ Now we look for $b$ : taking $(x,y)=(x_n,f(x_n))$ (because this point lies on the graph of $f$ and on the tangent), we get $$b=y-f'(x_n)x=f(x_n)-f'(x_n)x_n$$ so that the equation of the tangent becomes $$y=f'(x_n)x+f(x_n)-f'(x_n)x_n=f'(x_n)(x-x_n)+f(x_n).$$

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