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This question is in my book as an example, first thing I did was to put $y=0$ and find the value of $x$, which came out to be $7$ and by substituting this value in equation I got y=0.

When I differentiated this equation I got $1-[(2x-1)(x-7)]/[(x-2)(x-3)]^2$

And by substituting value of x in this equation tangent comes $1/400$

But, before proceeding further I decided to check form example if I am going correct.

And to my surprise, in the book before putting values of x in the differentiated equation was--

$[1-y(2x-5)]/[(x-2)(x-3)]$

Tried hard but can't figure out how and why they did it.

Any help from you will be appreciated. Thank You

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    is the equation $$y=\frac{x-7}{(x-2)(x-3)}$$? – 2017-01-08
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    Yes, I'm new here, sorry for inconvenience – 2017-01-08

2 Answers 2

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I don't well understand what is your question.

If you want the tangent line to the curve, the answer of Dr.Sonnhard is what you want. If your problem is to understand why the derivative is written as $[1-y(2x-5)]/[(x-2)(x-3)]$, note that: $$ y'=\frac{(x-2)(x-3)-(x-7)(2x-5)}{(x-2)^2(x-3)^2}= $$ $$ =\frac{(x-2)(x-3)}{(x-2)^2(x-3)^2}-\frac{(x-7)(2x-5)}{(x-2)^2(x-3)^2}= $$ $$ \frac{1}{(x-2)(x-3)}-\frac{x-7}{(x-2)(x-3)}\cdot\frac{2x-5}{(x-2)(x-3)}= $$ $$ \frac{1}{(x-2)(x-3)}-y\cdot\frac{2x-5}{(x-2)(x-3)}=\frac{1-y(2x-5)}{(x-2)(x-3)} $$

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    THANK YOU! Just did a silly mistake, lack of sleep probably be the reason. – 2017-01-08
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the first derivative is given by $${\frac {1}{ \left( x-2 \right) \left( x-3 \right) }}-{\frac {x-7}{ \left( x-2 \right) ^{2} \left( x-3 \right) }}-{\frac {x-7}{ \left( x- 2 \right) \left( x-3 \right) ^{2}}} $$ for $x=7$ we have $$\frac{1}{20}$$ thus our tangent line has the form $$y=\frac{1}{20}x+n$$ one can calculate $n$ by inserting $$x=7,y=0$$