# Thread:

1. ## Just another trivial problem

Hello, i hope you can help me with this trivial problem(i hope)

Determine the the equation for the tangent to the curve y=tan x/x

at the given point of ((pi/4),(4/pi))

Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

Thank you, Jones

2. Originally Posted by Jones
Hello, i hope you can help me with this trivial problem(i hope)

Determine the the equation for the tangent to the curve y=tan x/x

at the given point of ((pi/4),(4/pi))

Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

Thank you, Jones
The equation $y-y_1=k(x-x_1)$ is the equation of the line through
the point $(x_1,y_1)$ with slope $k$. For it to be the equation
of the tangent to the curve at the given point $k$ must be the slope
of the curve at the point. That is:

$<br /> k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}<br />$,

and $x_1=pi/4,\ y_1=4/pi$.

RonL

3. Don't we need to differentiate the function first?

4. Originally Posted by Jones
Don't we need to differentiate the function first?
That appears to be what 5th line of my earlier post says

RonL

5. Originally Posted by CaptainBlack
For it to be the equation
of the tangent to the curve at the given point $k$ must be the slope
of the curve at the point. That is:

$<br /> k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}<br />$,

$<br /> \frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}<br />$.

Therefore

$<br /> k=\frac{8}{\pi}-\frac{16}{\pi^2}\approx 0.92534<br />$

RonL

6. Originally Posted by CaptainBlack
$<br /> <br /> -\frac{\tan(x)}{x^2}<br />$.
where did this come from?

Im a little slow you see, almost dumb. therefor you have to be very legible

7. Originally Posted by Jones
where did this come from?

Im a little slow you see, almost dumb. therefor you have to be very legible
From the product rule for differentiation.

$<br /> \frac{d}{dx} f(x)g(x)= f'(x)g(x)+f(x)g'(x)<br />$

and here $g(x)=1/x$ and $f(x)=\tan(x)$ so:

$<br /> \frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}<br />$

I could have used the quotient rule, but since it is equivalent to this
use of the product rule I cant be bothered to remember it (as it is
redundant).

RonL

8. Originally Posted by CaptainBlack
I could have used the quotient rule, but since it is equivalent to this use of the product rule I cant be bothered to remember it (as it is redundant).
RonL
I'm glad I'm not the only one with that "hang-up."

-Dan

9. Originally Posted by topsquark
I'm glad I'm not the only one with that "hang-up."

-Dan
I'm not sure it is a "hang-up" I find it difficult to see why time and
limited human memory resources are wasted on teaching the quotient
rule.

It seems to me that the teaching of the q-rule in the anomaly -
one of those distrubing features of this reality which makes one suspect
that one has slipped into a parallel universe without having noticed.

RonL

10. Originally Posted by CaptainBlack
I'm not sure it is a "hang-up" I find it difficult to see why time and
limited human memory resources are wasted on teaching the quotient
rule.

It seems to me that the teaching of the q-rule in the anomaly -
one of those distrubing features of this reality which makes one suspect
that one has slipped into a parallel universe without having noticed.

RonL
I think I live in a parallel universe. (If I only had a "brane!" sings the scarecrow.)

-Dan
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