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Consider the graph defined by

$$y = \dfrac{x - 3}{x - 5}$$

(a) Use the definition of the derivative to find the slope of the tangent line to the graph at the point $(6, 3)$.

(b) Find an equation of the tangent line to the graph at the point $(6, 3)$.

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The definition of the derivative , which gives the slope is $$y_0'=\lim\limits_{x\to x_0}\dfrac{y-y_0}{x-x_0}$$ The equation of the tangent line is $$y-y_0=y'_0(x-x_0)$$ Can you now continue?

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  • $\begingroup$ I thank you for your help.Using this definition I find the slope of the tangent line to be 0.4 and the resulting equation to be y = 0.4x + 0.6 $\endgroup$ – Tangles May 25 '13 at 13:43
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$$\dfrac {dy}{dx}=\dfrac{(x-5)\dfrac {d}{dx}(x-3)-(x-3)\dfrac {d}{dx}(x-5)}{(x-5)^2}$$

$$\dfrac {dy}{dx}=\dfrac{(x-5)-(x-3)}{(x-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(x-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=\dfrac{-2}{(6-5)^2}$$ $$\dfrac {dy}{dx}_{(6,3)}=-2$$

slope of tangent at $(6,3)$=$-2$

b)

equation of tangent : $y-y_0=m(x-x_0)$

(where $(x_0,y_0)$ is a point on line and m is slope)

$$y-3=-2(x-6)$$ $$2x+y=15$$

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  • $\begingroup$ Thankyou so much :) Rockin answer! $\endgroup$ – Tangles May 25 '13 at 13:49
  • $\begingroup$ @Tangles my pleasure.Nothing rocking Just an starter question. $\endgroup$ – iostream007 May 25 '13 at 13:51
  • $\begingroup$ what can I say.... I'm a freshie in the world of calculus ;P $\endgroup$ – Tangles May 25 '13 at 13:53
  • $\begingroup$ keep continue buddy $\endgroup$ – iostream007 May 25 '13 at 13:54
  • $\begingroup$ This can also be greatly simplified by the observation that $\frac{x-3}{x-5}=1+\frac{2}{x-5}$. $\endgroup$ – Ataraxia May 25 '13 at 13:59

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