A-level Mathematics/OCR/C4/Advanced Differentiation

In Core 4 we will discuss the derivatives of trigonometric functions, implicit differentiation and parametric differentiation.

Derivatives of Trigonometric Functions
In order to find the derivatives of any of the six trigonometric function you only need to know the derivatives of the sine and cosine functions. One important thing to note is that when working with derivatives your angles need to be in radians.
 * If $$y = \sin kx,\,$$ then $$\frac{dy}{dx} = k\cos kx$$.
 * If $$y = \cos kx,\,$$ then $$\frac{dy}{dx} = -k\sin kx$$.
 * If $$y = \tan kx,\,$$ then $$\frac{dy}{dx} = k\sec^2 kx$$.
 * If $$y = \operatorname{cosec}\ kx,\,$$ then $$\frac{dy}{dx} = -k\operatorname{cosec}\ kx\cot kx$$.
 * If $$y = \sec kx,\,$$ then $$\frac{dy}{dx} = k\tan kx\sec kx$$.
 * If $$y = \cot kx,\,$$ then $$\frac{dy}{dx} = -k\operatorname{cosec}^2\ kx$$.

Implicit Differentiation
In C3, we learned how to differentiate explicit equations in the form of y=f(x), immediately obtaining the result of dy/dx also as a function of x. In C4, although the techniques are still used, you will need to know how to differentiate implicit functions. That is to say, differentiate both sides of an equation f(x,y)=g(x,y).

With implicit differentiation, all of the normal rules apply - however, when differentiating a term in y with respect to x, the result must have an additional dy/dx term as a result of the chain rule:

$$\frac{d}{dx}(f(y))=f'(y)\cdot\frac{dy}{dx}$$

So, rules like the familiar "multiply by the power, reduce the power" still apply in essence, although the dy/dx must be included. When differentiating implicitly, one still differentiates each term in turn. For example, a question might ask you to differentiate both sides, with respect to x, of the equation:

$$\sec (y) + x^2 = 3x + 2y + 1\;$$

Differentiating each term with respect to x, we get:

$$\sec (y)tan(y)\cdot\frac{dy}{dx} + 2x = 3 + 2\cdot\frac{dy}{dx}$$

In general, a question will ask for a function of x and y which gives the gradient at a particular point. So, we need to rearrange the expression for dy/dx:

$$\sec (y)tan(y)\cdot\frac{dy}{dx}-2\cdot\frac{dy}{dx} = 3 - 2x$$

$$\frac{dy}{dx}(\sec(y)tan(y)-2) = 3 - 2x$$

$$\frac{dy}{dx} = \frac{3-2x}{sec(y)tan(y)-2}

$$

From this point on, you may have to substitute values for x and y in order to compute the value of dy/dx at a particular location on the curve. Substitutions can be performed before rearrangement (to aid in factorisation) if the value is required, and not the expression. In some cases, a question may ask you to find the co-ordinates of points on the curve which satisfy a particular value of dy/dx. In this case, you will need to find the value of x where the differential equation is satisfied, or an equation that relates x and y, then substitute this into the original equation of the curve. This often results in multiple points which satisfy the given conditions.

Take (as a simple example) the curve:

$$x^3+y^3 = 5x\;$$

For which we are required by some arbitrary question to find the points at which dy/dx=0.

Although we could quite easily rearrange this to give an explicit function, for the case of demonstration, we will use implicit differentiation. Differentiating with respect to x yields:

$$3x^2 + 3y^2\frac{dy}{dx} = 5$$

$$\frac{dy}{dx}= \frac{5-3x^2}{3y^2} = 0$$

$$5-3x^2=0 \;$$

$$x = \pm\sqrt{\frac{5}{3}} \;$$

So, we simply substitute that value into the original implicit equation, and solve for y (twice). In some cases, you may find that x is a function of y given the conditions, and not simply a value. In this case, it is just a matter of solving a pair of simultaneous equations.

So what of the term 'xy'? Well, we simply apply the product rule:

$$(uv)'=vu'+uv'\;$$ where u and v are functions of x or y in this case.

$$(xy)'=y\cdot 1+x\cdot \frac{dy}{dx}$$

Similarly:

$$(x^2y)' = y\cdot 2x + x^2\cdot \frac{dy}{dx}$$

Related rates of change and differential equations
In this rather interesting topic, you will be looking at investigating how rates of change can affect one another, and how to write differential equations given a description of a situation. Again, C4 has merely found a way to complicate an otherwise easy process - the main difficulty is not concerned with mathematics as such. It is in deciphering a paragraph into a single line. Often, problems require knowledge of the use of the chain rule in differentiation.

For C4, we are mostly interested in:
 * Leaky tanks
 * Stains or other circular shapes

Let us imagine a situation in which there is a tank with volume, V. Its contents are leaking out at a rate proportional to the amount contained within the tank. So, we can write this as:

$$\frac{dV}{dt}=-kV$$

The negative sign being there because the tank is leaking. Next, let us say that in addition to this, there is the mandatory tap supplying the tank with fluid at a constant rate of 5$$m^3t^{-1}$$.

$$\frac{dV}{dt}= 5-kV$$

So what is the rate of change of height with respect to time? First, we need to examine the equation relating height to volume. Assuming that the tank is a cylinder of uniform cross section, we can say that:

$$V=\pi r^2h\;$$, where r is the radius of the tank at any point, and h is the height of the volume contained.

Differentiating with respect to h gives:

$$\frac{dV}{dh}=\pi r^2$$

$$\frac{dh}{dV}=\frac{1}{\pi r^2}$$

And so:

$$\frac{dh}{dt}=\frac{dh}{dV}\cdot\frac{dV}{dt}=\frac{1}{\pi r^2}(5-kV)$$

Where k is a constant of proportionality, usually to be figured out by a boundary condition.

Parametric differentiation
In parametric geometery, differentiation works as normal. One can either convert the curve into cartesian form, or use the result of the chain rule to find dy/dx:

$$\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}$$