Implicit Differentiation
Implicit differentiation is just a method for finding a derivative when
is not expressed explicitly as a function of . We use the Chain Rule.
Example: Find if
Solution: We differentiate both sides with respect to
.
then moves terms around to solve for
To find a second derivative we differentiate with respect to
again, in this case, using the quotient rule.
and substitute the quantity we found earlier
and substitute from the original equation
This last bit doesn't always turn out so nicely but it frequently does with standard functions.
STEP 2011, Math I, #1: (i) Show that the gradient of the curve
where , is
.
The point
lies on both the straight line and the curve , where . Given that, at this point, the line
and the curve have the same gradient, show that
.
Show further that either or .
(ii) Show that if the straight line
, where , is a normal to the curve , then .
STEP 2006, Math II, #7: An ellipse has equation
. Show that the equation of the tangent at the point is
The point A has coordinates
, where and are positive. The point E has coordinates and the point P has coordinates
where . The line through E parallel to AP meets the line at the point Q. Show that the line PQ is tangent to the
above ellipse at the point given by
.
Determine by means of sketches, or otherwise, whether this result holds also for
and .
STEP 2009, Math I, #2: A curve has the equation
where
and are positive constants. Show that the tangent to the curve at the point is
.
In the case
and , show that the -coordinate of the points where the tangent meets the curve satisfy
.
Hence find positive integers
, , and such that
.
STEP 2010, Math II, #1: Let P be a given point on a given curve C. The osculating circle to C at P is defined to be the circle that satisfies
the following two conditions at P: it touches C; and the rate of change of its gradient is equal to the rate of change of the gradient at C.
Find the centre and radius of the osculating circle to the curve
at the point on the curve with -coordinate .
STEP 2008, Math I, #8: (i) The gradient
of a curve at a point satisfies
(*)
By differentiating (*) with respect to
, show that either or .
Hence show that the curve is either a straight line of the form
, where , or the parabola .
(ii) The gradient
of a curve at a point satisfies
Show that the curve is either a straight line, the form of which you should specify, or a circle, the equation of which you should determine.