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Using first principles, differentiate $$f(x)=(x)^{\frac{3}{4}} $$

So$$\lim_{h\to 0}\frac{f(x+h) -f(x)}{h} = \lim_{h\to 0}\frac{(x+h)^{\frac{3}{4}} -(x)^{\frac{3}{4}}}{h} $$

I cant seem to solve this question. I'm supposed to get the output $\frac{3}{4}x^{-\frac{1}{4}} $

nmasanta
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Peachh Salts
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1 Answers1

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The key to solving this problem is to apply the binomial series to $(x+h)^\frac{3}{4}$. $$(x+h)^\frac{3}{4}=x^\frac{3}{4}(1+\frac{h}{x})^\frac{3}{4}=x^\frac{3}{4}(1+\frac{3h}{4x}+...).$$ Then $(x+h)^\frac{3}{4}-x^\frac{3}{4}=\frac{3h}{4}x^{-\frac{1}{4}}+...$

and $\frac{(x+h)^\frac{3}{4}-x^\frac{3}{4}}{h}=\frac{3}{4}x^{-\frac{1}{4}}.$

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    When you expand $ (1+\frac{h}{x})^{\frac{3}{4}} $, it seems like you are using a Taylor Series expansion, which I'm not sure counts as first principles. I feel like there should be a way just using algebraic manipulations in the difference quotient to arrive at the desired derivative. – poopstraw Sep 10 '19 at 08:45
  • For a school lesson I think that first principles simply implies letting $h$ tend to zero using the techniques they already know. I don't think it would help this student to do any other method –  Sep 10 '19 at 08:49
  • Binomial expansions with non integer powers are difficult to understand or motivate without calculus. I agree with @poopstraw, there is a standard algebraic manipulation using conjugates (in this case, using difference of squares twice) i.e. multiply the numerator and denominator by $((x+h)^{\frac{3}{4}}+x^{\frac{3}{4}})((x+h)^{\frac{3}{2}}+x^{\frac{3}{2}})$ – Ninad Munshi Sep 10 '19 at 08:55
  • The binomial series is a standard school topic which is expected in questions like this one. –  Sep 10 '19 at 09:04
  • @S.Dolan I disagree, this is not a standard high school topic at all. In order to prove the formula works for non integer powers, one needs calculus, so it would be very strange to be using it before having that tool. Not to mention, that binomial series is only valid for $|x|<1$. – Ninad Munshi Sep 10 '19 at 16:57