Differentiation question

Hi, I hope anyone reading this is having a great day. Can someone explain to me this question, it would very much appreciated. Specially the third step when they did x^1/6. Initially tried using the chain rule where I set something equal to u but it didn’t work.

(edited 2 years ago)

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Reply 1

mqb2766

Original post by Eris13696

if you have something like
(3x - 5x^2)^2
That is equal to
x^2(3-5x)^2

They've done the reverse and its raised to the power 6 so the multiplier of 1/x actually multiplies the terms in each of the 6 brackets by (1/x^(1/6))

Is that the part you're asking about? The derivative does use the chain rule, so if you have something different, upload what you tried?

(edited 2 years ago)

Reply 2

DFranklin

It has to be said, the method they've used is utterly bizarre. Far simpler to simply differentiate as the product

x(5-3x^2)^{-6}

, and then put everything over a common denominator of

(5-3x^2)^7

Reply 3

Eris13696

Original post by DFranklin

It has to be said, the method they've used is utterly bizarre. Far simpler to simply differentiate as the product

x(5-3x^2)^{-6}

, and then put everything over a common denominator of

(5-3x^2)^7

Thank you, if it doesn’t trouble you much, I’m still not introduced to the product rule, it’s in pure maths 2 and this is a pure math 1 question, so far I just know second derivatives and the chain rule, is there any other way to solve it?

Reply 4

davros

Original post by Eris13696

Are you sure you haven't covered the product rule? Every presentation of calculus I've ever seen has done product rule then quotient rule then chain rule. Is this the current A level course?

There is *another* way of doing this, but the normal way to differentiate a quotient is either to use the quotient rule itself, or as @DFranklin suggests, rewrite as a product and then use the product rule...

Reply 5

Eris13696

Original post by davros

Yeah I’m still doing the AS part. The product rule is in the A level part of the book which I haven’t reached yet, guess for now I will skip this question until I reach A level part.

I’m doing the international A levels maybe that’s why it’s in a different sequence :0? I think that’s why they did this very complicated way of solving the question because they didn’t introduce the rules yet. I’m grateful for the replies, I will keep it in mind when I do the A level book soon!!

Reply 6

davros

Original post by Eris13696

Have you covered the natural logarithm ln yet?

If you write u = dy/dx then you want to evaluate du/dx.

Taking logs of the original expression gives us

\ln u = \ln x - 6ln(5 - 3x^2)

Differentiating with respect to x and using the chain rule now gives

Unparseable latex formula:

\displaystyle \dfrac{1}{u} \times \dfrac{du}{dx} = \dfrac{1}{x} + \dfrac{36x}{5 - 3x^2}}

\displaystyle \dfrac{du}{dx} = \dfrac{x}{(5- 3x^2)^6} \times \left(\dfrac{1}{x} + \dfrac{36x}{5 - 3x^2} \right) = \dfrac{1}{(5- 3x^2)^6} + \dfrac{36x^2}{(5 - 3x^2)^7}

\displaystyle = \dfrac{5 - 3x^2}{(5-3x^2)^7} + \dfrac{36x^2}{(5 - 3x^2)^7}

\displaystyle = \dfrac{5 + 33x^2}{(5 - 3x^2)^7}

Reply 7

DFranklin

Original post by davros

There is *another* way of doing this, but the normal way to differentiate a quotient is either to use the quotient rule itself, or as @DFranklin suggests, rewrite as a product and then use the product rule...

Given the extremely strange "approved method" I think it's likely the OP is right and they only have the chain rule at this point (or at least, the approved solution assumes they only have the chain rule)...
Nice log solution. For completeness (and because I think the point here is to say *why* we think the given solution is unnecessarily painful):

\displaystyle \dfrac{d}{dx} \dfrac{x}{(5-3x^2)^6} = \dfrac{d}{dx} x (5-3x^2)^{-6} = (5-3x^2)^{-6} \dfrac{d}{dx} x + x \dfrac{d}{dx} (5-3x^2)^{-6}

\displaystyle = (5-3x^2)^{-6} +x (-6x)(-6) (5-3x^2)^{-7} = \dfrac{1}{(5-3x^2)^6} + \dfrac{36x^2}{(5-3x^2)^7}

and then tidy up to the desired result as with the previous solution.

Reply 8

davros

Original post by DFranklin

\displaystyle \dfrac{d}{dx} \dfrac{x}{(5-3x^2)^6} = \dfrac{d}{dx} x (5-3x^2)^{-6} = (5-3x^2)^{-6} \dfrac{d}{dx} x + x \dfrac{d}{dx} (5-3x^2)^{-6}

\displaystyle = (5-3x^2)^{-6} +x (-6x)(-6) (5-3x^2)^{-7} = \dfrac{1}{(5-3x^2)^6} + \dfrac{36x^2}{(5-3x^2)^7}

and then tidy up to the desired result as with the previous solution.

That's probably how I would do it, although there's not much between it and a direct application of the quotient rule (which is of course equivalent).

I'm struggling to understand how any "sensible" course teaches the chain rule before the product rule, but then my calculus was self-taught (from 'Calculus Made Simple') before I'd even seen an A level textbook, so I find the modern obsession with modulariizing everything in sight is utterly baffling :biggrin: