9709. P3. Differentiation

What do we remember from last year’s differentiation work? Let’s take a look at last year’s syllabus extract to remind us:

This year we will learn some further ways of differentiating more complicated functions, including those transcendental functions that we have recently learned.

Product Rule

As is often the case in A-level mathematics, we need primarily to be able to use the product rule, but we should also have some understanding of the derivation of the product rule from first principles. It is derived below:

Let P(x) = f(x)g(x).

Then P'(x) = \lim_{\delta x \to 0} \frac{P(x+\delta x)-P(x)}{\delta x}

= \lim_{\delta x \to 0} \frac{f(x+ \delta x)g(x+ \delta x) - f(x)g(x)}{\delta x}

Now we will do a *strange* step, which you would not be expected to have anticipated yourself, but we do it because it works. In mathematics introducing something extra (e.g. multiplying the numerator and denominator of a fraction by the same thing) can often help us, and sometimes it is by trial and error that mathematicians find those things that we should introduce. So we are going to add and subtract f(x+𝛿x)g(x) to the numerator of the fraction, as this will allows us to factorise the numerator into a useful form.

So, \lim_{\delta x \to 0}({ \frac{f(x+ \delta x)g(x+ \delta x) - f(x)g(x)}{\delta x} })

= \lim_{\delta x \to 0}({ \frac{f(x+ \delta x)g(x+ \delta x)-f(x + \delta x)g(x) + f(x + \delta x)g(x) - f(x)g(x)}{\delta x} })

= \lim_{\delta x \to 0} (\frac {f(x+\delta x)[g(x+\delta x)-g(x)]+g(x)[f(x+\delta x)-f(x)]} {\delta x} )

= \lim_{\delta x \to 0}( f(x+\delta x)[\frac {g(x+\delta x)-g(x)}{\delta x}]+g(x)[\frac{f(x+ \delta x)-f(x)}{\delta x} ])

= \lim_{\delta x \to 0}f(x+\delta x).\lim_{\delta x \to 0}\frac {g(x+\delta x)-g(x)}{\delta x}+\lim_{\delta x \to 0}g(x).\lim_{\delta x \to 0}\frac{f(x+ \delta x)-f(x)}{\delta x}

=f(x)g'(x) + g(x)f'(x).

You should be able to follow the logic of this argument, but would not be required to demonstrate it in an examination.

Worked Examples

1.) y=(2x+3)(x2-5). Using the product rule, find \frac{dy}{dx}.

2.) Differentiate y=20x(x-1)6

3.) If f(x)=x^2\sqrt{3x-1}, then find f'(x)

4.) Given that y = (x+1)(x-2)2,

  • find \frac{dy}{dx};
  • find any stationary points and determine their nature;
  • sketch the curve.



Quotient Rule

We can utilise the work we did with the product rule to demonstrate the quotient rule.

Let Q(x)=\frac{f(x)}{g(x)}.

We can rewrite this as f(x)=Q(x)g(X)

Applying the product rule tells us that f'(x) = Q'(x)g(x) + Q(x)g'(x).

Solving this for Q'(x), we get Q'(x) = \frac{f'(x) - Q(x)g'(x)}{g(x)}

= \frac{f'(x) - \frac{f(x)}{g(x)}.g'(x)}{g(x)}

= \frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^2}

Worked Examples

1.) Given that \frac{3x+1}{x-2}, use the quotient rule to find \frac{dy}{dx}

2.) Given that y= \frac{x^2+1}{3x-1}, find \frac{dy}{dx} using the quotient rule.

3.) Given that y=\frac{x-3}{x-4}:

  • find \frac{dy}{dx}
  • find the equation of the tangent to the curve at the point (6,15)
  • find the equation of the normal to the curve at the point (5,2)
  • use the answer to the first part above to deduce that the curve has no stationary points, and sketch its graph.



Differentiating ex

By our original definition of derivatives, if f(x)=ex, then:

f'(x) = lim_{\delta x \to 0} \frac{e^{x+\delta x} - e^x}{\delta x}

=lim_{\delta x \to 0} \frac{e^x(e^{\delta x}-1)}{\delta x}

=e^x.lim_{\delta x \to 0}\frac{e^{\delta x}-1}{\delta x}

We haven’t yet learned methods that would allow us to calculate lim_{\delta x \to 0}\frac{e^{\delta x}-1}{\delta x}, but if we try some values close to zero, we see the following:

\frac{e^{\delta x}-1}{\delta x}0.95160.99500.999511.00051.00501.0517

So we can see that lim_{\delta x \to 0}\frac{e^{\delta x}-1}{\delta x} = 1, so the derivative of ex ix ex.1, or simply ex.



(ef(x))’ = ef(x).f'(x)

Worked Examples

1.) Differentiate y = e5x

2.) Find the derivative of the function f(x) = x2ex

3.) Find the derivative of the function f(x) = (e^x+1)^{\frac{5}{2}}

4.) Differentiate f(x) = e3x+1

5.) Find \frac{dy}{dx} if f(x)=xe^{x^2-1}



Differentiating ln(x)

We use the fact that ln(x) is the inverse of ex to help us find its derivative.

y = ln(x). Treating both sides of the equation as powers of y gives us:

ey=x. Differentiating both sides with respect to y gives us:

ey=\frac{dx}{dy}. Reciprocating both sides then gives us

\frac{1}{e^y} = \frac{dy}{dx}.

So \frac{dy}{dx} = \frac{1}{x}
We can apply the chain rule to see that (ln(f(x))' = \frac{f'(x)}{f(x)}

Worked Examples

1.) Find the derivative of the function f(x) = xln(x)

2.) Find the derivative of the function f(x) = ln\sqrt[5]{x^3}

3.) Differentiate f(x) = (lnx + x)2

4.) Differentiate f(x) = ln(x2+3x+1)



Differentiating Trigonometric Functions

Again using our definition of a derivative with sine x, we get:

f(x) = sin(x), so:

f'(x) = lim_{\delta x \to 0} \frac{sin(x + \delta x)-sin(x)}{\delta x}

= lim_{\delta x \to 0} \frac{sin(x)cos(\delta x)+cos(x)sin(\delta x) - sin(x)}{\delta x}

= lim_{\delta x \to 0} \frac{sin(x)cos(\delta x) - sin(x)}{\delta x} + \frac{cos(x)sin(\delta x)}{\delta x}

= lim_{\delta x \to 0} sinx(\frac{cos(\delta x)-1}{\delta x}+cosx \frac{sin(\delta x)}{\delta x})

= sin(x).lim_{\delta x \to 0}\frac{cos(\delta x) -1}{\delta x} + cos(x).lim_{\delta x \to 0}\frac{sin(\delta x)}{\delta x}

For the final step, we need to convince ourselves geometrically (thinking about the unit circle may persuade us a bit) that the following statments are true:

lim_{\delta x \to 0} \frac{sin(\delta x)}{\delta x}=1 and lim_{\delta x \to 0} \frac{cos(\delta x)-1}{\delta x}=0 .

Putting these into the above gives us that f'(x) = sin(x).0 + cos(x).1 = cos x.

To find the derivative of cosine x, we notice that cos(x) = sin(90-x), so (cos(x))’ = (sin(90-x))’ = cos(90-x).(-1) = -cos(90-x) = -sin(x).

We can find the derivative of tan(x) by using the quotient rule.

In general, when using trigonometry in calculus problems, we will always use radians unless explicitly directed otherwise.

Summary of Formulae
(sin(x))’ = cosx

(cos(x))’ = -sinx

(tan(x)’ = sec2x = 1 + tan2x

(sin(f(x))’ = cos(f(x)).f'(x)

(cos(f(x))’ = -sin(f(x)).f'(x).

Worked Examples

1.) Find the derivative of the function f(x)=(sinx + cosx)2

2.) Differentiate xsinx

3.) Find the derivative of the function f(x) = \frac{e^x}{cosx}

4.) Differentiate f(x) = tan (x2-3x+1)

5.) Find the derivative of y=\frac{tanx}{x}



Implicit Differentiation

If y is a function of x but cannot be conveniently made into the subject of the equation then we can use implicit differentiation.

After differentiating we rearrange the equation to make y’ the subject.

1.) Differentiate each of the following with respect to y:

  • y2
  • xy
  • 3x2y3
  • sin(y)

2.) The equation of a curve is given by y3+xy=2

  • Find an expression for \frac{dy}{dx} in terms of x and y
  • Hence find the gradient of the curve at (1,1) and the equation of the tangent to the curve at that point.

3.) Differentiate x3+y3=3xy with respect to x and hence find the coordinates of any stationary points.



Parametric Differentiation

Some complicated equations are better expressed by the use of a third variable, or a parameter.

  • e.g. x=2+cost;
  • y=t+sint.

Putting values of t in gives us the coordinate pairs in order to plot the graph of y against x.

We differentiate by calculating \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

Worked Examples

1.) A curve has parametric equations: x=t2 and y=2t:

  • Find \frac{dy}{dx} in terms of parameter t;
  • Find the equation of the tangent to the curve at the general point (t2,2t)
  • Find the equation of the tangent at the point where t=3.

2.) A curve has parametric equations x=4cos𝜽, y=3sin𝜽.

  • Find \frac{dy}{dx} at the point with parameter 𝜽
  • Find the equation of the normal at the general point (4cos𝜽,3sin𝜽)
  • Find the equation of the normal at the point where 𝜽=π/4
  • Find the coordinates of the point where 𝜽=π/4

3.) Find the stationary points of the curve with parametric equations x=2t+1, y=3t-t3.


Mixed Exercise

Exercise Answers

Mixed Exercise Answers

%d bloggers like this: