I was asked to differentiate $((1-3x)^5-x^2)^4$ with respect to x, I thought that I would have to apply chain rule to the internal function before moving on to the outside. However, the answer was $-4((1-3x)^5-x^2)^3(15(1-3x)^4+2x)$. How was this achieved so simply?

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    $\begingroup$ A quick check says the answer is clearly wrong... unless perhaps you misread the question? $\endgroup$ – Simply Beautiful Art Feb 15 '17 at 14:39
  • $\begingroup$ yes you can do it with the chain rule, like unwrapping a parcel $\endgroup$ – Cato Feb 15 '17 at 14:46
  • $\begingroup$ Sorry i forgot to add the second part of the answer $\endgroup$ – MetaKnight35 Feb 15 '17 at 14:47

The answer is most probably wrong.

The common method to do this type of problems is by applying the chain rule.

Let,$y=[(1-3x)^5-x^2]^4$ and let,$z=(1-3x)^5-x^2$

So,$$\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}\implies \frac{dy}{dx}=\frac{d}{dz}z^4\cdot[\frac{d}{dx}[(1-3x^5)-x^2]]=4z^3\cdot[\frac{d}{dx}[(1-3x^5)-x^2]]$$.

Now,you can continue as usual.(Don't forget to substitute the $z$ at last).

  • $\begingroup$ ok thanks! I think I get it now $\endgroup$ – MetaKnight35 Feb 15 '17 at 14:50
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    $\begingroup$ Typo: $z^5$ should be $z^4$, which changes your derivative from $5z^4$ to $4z^3$ $\endgroup$ – Giulio Feb 15 '17 at 14:50

The "official" answer seems to be correct. Awkward image of derivation follows:

image of wordpad text


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