Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2019 | Oct-Nov | (P2-9709/21) | Q#3

Question

A curve has equation Find the exact gradient of the curve at the point for which y = 4.

Solution

We are required to find the gradient of the curve at a point where y=4.

Gradient (slope) of the curve at the particular point is the derivative of equation of the  curve at that particular point.

Gradient (slope) of the curve at a particular point can be found by

substituting x-coordinates of that point in the expression for gradient of the curve; Therefore, first we need expression for gradient of the curve.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence  gradient of curve with respect to is: We are given that; Hence; We differentiate using Quotient Rule.

If and are functions of , and if , then; If , then; Let and , then; We differentiate and one by one.

First we differentiate . Rule for differentiation of is:  Rule for differentiation of is: For ;   Next, we differentiate . Rule for differentiation of is:  Rule for differentiation of is: For ;   Hence;   To find the gradient of the curve at point y=4, we need to substitute value of x at this  point in this expression.

We need to find x-coordinate of the point for which y-coordinate is given.

Corresponding values of x coordinate can be found by substituting value of y in  equation of the curve.     Taking anti-logarithm of both sides;  and are inverse functions. The composite function is an identity function, with domain the positive real numbers. Therefore;  We substitute in expression for gradient of the curve to find gradient of the  curve at the point where y=4.   for any       