The equation is the equation of the line throughOriginally Posted by Jones
the point with slope . For it to be the equation
of the tangent to the curve at the given point must be the slope
of the curve at the point. That is:
,
and .
RonL
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 is the equation of the line throughOriginally Posted by Jones
the point with slope . For it to be the equation
of the tangent to the curve at the given point must be the slope
of the curve at the point. That is:
,
and .
RonL
From the product rule for differentiation.Originally Posted by Jones
and here and so:
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 not sure it is a "hang-up" I find it difficult to see why time andOriginally Posted by topsquark
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