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Based on $f(x) = x^2$, we know $f'(x) = 2x$.

If $x = y^2$

Then $f(y) = (y^2)^2 = y^4$, and $f'(y) = 4y^3$.

My question is if we only know

  • $f'(x) = 2x$
  • $x = y^2$

How to get $f'(y) = 4y^3$ ? Any hint or formula will be helpful. Thanks

Mark Viola
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Hongbo Miao
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2 Answers2

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Your $f'$ is incorrect; it should be $2x$. And what are you looking for is called Chain rule: https://en.wikipedia.org/wiki/Chain_rule

Botond
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  • Oh that is what I need, I will fix my question too! Thanks a lot! – Hongbo Miao May 20 '20 at 15:59
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    @HongboMiao You're welcome :) The easy way to remember the formula is $\frac{\mathrm{d}f}{\mathrm{d}y}=\frac{\mathrm{d}f}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}y}$, i.e. you revert the cancellation of $\mathrm{d}x$. The mathematicians don't really like this explanation, because it's not 100% correct, but it's much easier to remember. – Botond May 20 '20 at 16:02
  • @HongboMiao Sorry, my formula was wrong, but I fixed it. – Botond May 20 '20 at 16:07
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This is simply the chain rule. Since $x(y) = y^2, x'(y) = 2y$ and $$ \frac{d}{dy} f(x(y)) = f'(x(y)) x'(y) = 2(y^2) 2y = 4y^3. $$

gt6989b
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