0
$\begingroup$

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 $(6, 3)$.

|cite|improve this question
$\endgroup$
2
$\begingroup$

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?

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

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)

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.