Basic question of differentiation

HKDSE 2020 M2 Q.9 (b), a 2-mark question:

Given $\displaystyle f(x)=\frac{(x+4)^3}{(x-4)^2}$ , find f''(x) .

How fast can you finish this part and obtain the correct answer, especially when you are under the pressure during the public examination?

3 minutes? (2/100 * total time allowed = 2/100 * 150 minutes)

I try to do it in different ways.

Method 1 Direct application of quotient rule

$\displaystyle f'(x)$

$\displaystyle =\frac{3(x-4)^2(x+4)^2-2(x+4)^3(x-4)}{(x-4)^4}$

$\displaystyle =\frac{(x-4)(x+4)^2(3(x-4)-2(x+4))}{(x-4)^4}$

$\displaystyle =\frac{(x+4)^2(x-20)}{(x-4)^3}$

$\displaystyle f''(x)$

$\displaystyle =\frac{(x-4)^3((x+4)^2+2(x+4)(x-20))-3(x+4)^2(x-20)(x-4)^2}{(x-4)^6}$

$\displaystyle =\frac{(x-4)(x^2+8x+16+2x^2-32x-160)-3(x-20)(x^2+8x+16)}{(x-4)^4}$

$\displaystyle =\frac{(x-4)(3x^2-24x-144)-3(x^3-20x^2+8x^2-160x+16x-320)}{(x-4)^4}$

$\displaystyle =\frac{3x^3-24x^2-12x^2+96x-144x+576-3x^3+36x^2+432x+960}{(x-4)^4}$

$\displaystyle =\frac{384x+1536}{(x-4)^4}$

$\displaystyle =\frac{384(x+4)}{(x-4)^4}$

I don’t think this is a good method, because we can solve it in a faster way, about 1 minute.

Method 2 Making the denominator to be a monomial

$\displaystyle y=\frac{(x+4)^3}{(x-4)^2}$

Let u=x-4 , then

$\displaystyle y=\frac{(u+8)^3}{u^2}=u+24+\frac{192}{u}+\frac{512}{u^2}$

$\displaystyle y'=1-\frac{192}{u^2}-\frac{1024}{u^3}$

$\displaystyle y''=\frac{384}{u^3}+\frac{3072}{u^4}=\frac{384(u+8)}{u^4}=\frac{384(x+4)}{(x-4)^4}$

Done.

Here, we should be careful that y'' above means $\displaystyle \frac{d^2y}{du^2}$ which may not be equal to $\displaystyle \frac{d^2y}{dx^2}$ as required in general.

But for u=x-4 , we do have $\displaystyle \frac{d^2y}{du^2}=\frac{d^2y}{dx^2}$ . Thank God!

To tackle the differentiation of rational functions, logarithmic differentiation may always help. But the trick had been out of syllabus of M1 since 2016; and may not be that emphasized in M2.

Let me pick up this method to solve.

Method 3 Logarithmic differentiation

$\displaystyle y=\frac{(u+8)^3}{u^2}$

$\displaystyle \ln y=3\ln(u+8)-2\ln u$

$\displaystyle \frac{1}{y}y'=\frac{3}{u+8}-\frac{2}{u}=\frac{3u-2u-16}{u(u+8)}=\frac{u-16}{u(u+8)}$

$\displaystyle y'=\frac{(u+8)^3}{u^2}\frac{u-16}{u(u+8)}=\frac{(u+8)^2(u-16)}{u^3}$

$\displaystyle \ln y'=2\ln(u+8)+\ln(u-16)-3\ln u$

$\displaystyle \frac{1}{y'} y''=\frac{2}{u+8}+\frac{1}{u-16}-\frac{3}{u}$

$\displaystyle \frac{1}{y'} y''=\frac{2(u-16)u+(u+8)u-3(u+8)(u-16)}{(u+8)(u-16)u}$

$\displaystyle \frac{1}{y'} y''=\frac{2u^2-32u+u^2+8u-3u^2+24u+384}{(u+8)(u-16)u}$

$\displaystyle \frac{1}{y'} y''=\frac{384}{(u+8)(u-16)u}$

$\displaystyle y''=\frac{(u+8)^2(u-16)}{u^3}\frac{384}{(u+8)(u-16)u}=\frac{384(u+8)}{u^4}=\frac{384(x+4)}{(x-4)^4}$

Urm, I still need . Well, just a way.

From day 1, I said I don’t like tedious computations like this kind of question. And I guess some students might use product rule directly, whatever, hope them obtained correct answer in this part and use the result to help getting 7 marks of the rest of the curve sketching and finding area bounded by the curve and axes problems.

Exercise
1. (HKALE 2005 Pure Mathematics Paper II Q.7) Sketch the graph of

$\displaystyle y=-x+|x|\sqrt{\frac{x}{x+2}}$

for x 0 .

2. Let u=g(x) , what is the relation between $\displaystyle \frac{d^2y}{dx^2}$ and $\displaystyle \frac{d^2y}{du^2}$ ? (Differentiability assumed)

Basic question of differentiation

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