# 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

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