A Level Maths: Product Rule

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Let us consider the following expression which we want to differentiate y=x^2 \sqrt{1+x}

You can see that there are two functions which are being multiplied together.

The first function is that g(x)=u^2 \text { and } f(x)=\sqrt{1+x^2}

In order to differentiate such expressions we need to use the Product Rule which has a very simple formula: 

\text { If } y=u \times v \text {, then } \frac{\mathrm{d} y}{\mathrm{~d} x}=u \frac{\mathrm{d} v}{\mathrm{~d} x}+v \frac{\mathrm{d} u}{\mathrm{~d} x}

Applying this formula will give us the following: 

\begin{aligned} & \quad y=x^2 \sqrt{1+x} \\ & u=x^2 \quad v=(1+x)^{\frac{1}{2}} \\ & \frac{d u}{d x}=2 x \quad \frac{d v}{d x}=\frac{1}{2}(1+x)^{-\frac{1}{2}} \\ & \therefore \frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x} \\ & =\frac{1}{2} x^2(1+x)^{-\frac{1}{2}}+(1+x)^{\frac{1}{2}}(2 x) \end{aligned}

As for obtaining an answer, we have. We have successfully differentiated our original expression. 

Now with this question you can take your answer a step further and this involves a level of factorisation with the use of negative indices. It is quite often within A Level Maths Product Rule Questions you are asked to give your answer in terms of factorised form. The main benefit of this is that it allows you to be able to determine turning points very quickly.

For instance, with the above question we have:

\frac{d y}{d x}=\frac{1}{2} x^2(1+x)^{-\frac{1}{2}}+2 x(1+x)^{\frac{1}{2}}

This can be made simpler by factorising. When factorising it is best to take out any negative fractional powers as the common term as you will see in the fully worked solution below: 

\begin{aligned} & =x(1+x)^{-\frac{1}{2}}\left[\frac{1}{2} x+2(1+x)\right] \\ & =x(1+x)^{-\frac{1}{2}}\left[\frac{x+4(1+x)}{2}\right] \\ & =x(1+x)^{-\frac{1}{2}}\left[\frac{5 x+4}{2}\right] \\ & \therefore \frac{d y}{d x}=\frac{1}{2} x(1+x)^{-\frac{1}{2}}(5 x+4) \end{aligned}


To answer part a) you need to recognise that you have a function multiplied by another function and so the technique that is needed is the product rule for A level maths. 

It is also important to note that you are given the answer. But you need to derive the answer. As a part of your examination technique, if you were not able to obtain the desired answer that was provided, then you would use this answer in part b) and this would be acceptable. 

Differentiating part a) gives:

\begin{aligned} & y=x^3(1+3 x)^{\frac{1}{3}} \\ & \begin{aligned} u=x^3 \quad v & =(1+3 x)^{\frac{1}{3}} \\ \frac{d u}{d x}=3 x^2 \quad \frac{d v}{d x} & =\frac{1}{3}(1+3 x)^{-\frac{1}{3}}(3) \\ & =(1+3 x)^{-\frac{1}{3}} \\ \frac{d y}{d x} & =u \frac{d v}{d x}+v \frac{d u}{d x} \\ & =x^3(1+3 x)^{-\frac{1}{3}}+3 x^2(1+3 x)^{\frac{1}{3}} \end{aligned} \end{aligned}

One thing that is important is that quite often, even though you are using the product rule you are also having to use the chain rule and this is very common when doing A Level maths. 

To factorise this take out any common terms. You will notice that x^2 as is (1+3 x)^{-\frac{1}{3}}:

\begin{aligned} & =x^2(1+3 x)^{-\frac{1}{3}}[x+3(1+3 x)] \\ \therefore \frac{d y}{d x} & =x^2(1+3 x)^{-\frac{1}{3}}(10 x+3) \end{aligned}

You will find that taking any negative fractional powers out as a common term is much easier than using a positive fractional power. 

Part b) is asking us to determine the turning points of the curve and this can be done as follows: 

\frac{d y}{d x}=0\begin{gathered} x^2(1+3 x)^{-\frac{1}{3}}(10 x+3)=0 \\ x^2=0 \rightarrow x=0 \\ 1+3 x=0 \rightarrow x=-\frac{1}{3} \\ 10 x+3=0 \rightarrow x=-\frac{3}{10} \end{gathered}


We have the x coordinates we simply need to determine the y coordinates by using the original expression for y:

\begin{aligned} & x=0 \quad y=0 \quad \rightarrow(0,0) \\ & x=-\frac{1}{3} \quad y=0 \rightarrow\left(-\frac{1}{3}, 0\right) \\ & x=-\frac{3}{10} \quad y=-0.0125 \rightarrow\left(-\frac{3}{10},-0.0125\right) \end{aligned}

And these are the coordinates of the turning points. 

We will look at a few exam style questions where knowledge of other areas of pure maths at A Level has been assumed. These questions will involve the use of the product rule and the chain rule as well. 

Exam Question 1

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The full solution you will find below. Try the question on your own before looking at the solution provided: 

\begin{aligned} & y=x(x-1)^{\frac{1}{2}} \\ & u=x \quad v=(x-1)^{\frac{1}{2}} \\ & \frac{d u}{d x}=1 \quad \frac{d v}{d x}=\frac{1}{2}(x-1)^{-\frac{1}{2}} \\ & \frac{d y}{d x}=\frac{1}{2} x(x-1)^{-\frac{1}{2}}+(x-1)^{\frac{1}{2}} \end{aligned}

What has been done so far is that the gradient function has been obtained as has the value of the gradient at x = 5. The y value has also been found. These are now going to help us determine the equation of the normal. Remember that the gradient of the normal is the negative reciprocal of the gradient of the tangent. 

\begin{gathered} \text { normal gradient }=-\frac{4}{13} \\ y=-\frac{4}{13} x+c \quad(5,10) \\ 10=-\frac{20}{13}+c \\ c=\frac{150}{13} \\ y=\frac{-4}{13} x+\frac{150}{13} \end{gathered}


Exam Question 2 

For part a) we have the following: 

For this question it has been taken that you are aware of how to differentiate e^x and unless a question asks, there is no need to factorise your answer. This has been for neatness and nothing more. 

\begin{aligned} y & =2 x e^x \\ u & =2 x \quad v=e^x \\ \frac{d u}{d x} & =2 \quad \frac{d v}{d x}=e^x \\ \therefore \frac{d y}{d x} & =u \frac{d v}{d x}+v \frac{d u}{d x} \\ & =2 x e^x+2 e^x \\ & =2\left(x e^x+e^x\right) \end{aligned}

For part b) we have the following: 

\begin{aligned} y=3 x^2 \ln 2 x \\ u=3 x^2 \quad v=\ln 2 x \\ \frac{d u}{d x}=6 x \quad \frac{d v}{d x}=\frac{1}{x} \\ \frac{d y}{d x}=\frac{u d v}{d x}+v \frac{d u}{d x} \\ =3 x^2\left(\frac{1}{x}\right)+\ln 2 x(6 x) \\ =6 x \ln 2 x+3 x \end{aligned}

For the derivative of ln2x, this has made use of a standard result, for which there are many. 

Should you forget any of the standard results then you can always use the chain rule. Below is the working for the derivative of y = ln2x:

\begin{aligned} y & =\ln 2 x \\ u & =2 x \quad y=\ln u \\ \frac{d u}{d x} & =2 \quad \frac{d y}{d u}=\frac{1}{u} \\ \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =\frac{1}{u} \times 2 \\ & =\frac{1}{2 x} \times 2 \\ \therefore \frac{d y}{d x} & =\frac{1}{x} \end{aligned}

Differentiation is a very large area of A Level Maths and it is something that you must be able to do and to be able to do well. Future articles will look at other areas of calculus and past exam style questions which will show you how to obtain the maximum marks. 

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