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I'm here with a maths problem. I'm trying to find the derivative of an expression but when I calculate it I get different answers depending on if I use the power rule or definition of a derivative. The problem is: $$\sqrt[3]{x}$$

When I use the power rule I get $\frac{1}{3\sqrt[3]{x^2}}$

But I don't get that answer when I use the definition of a derivative.

$$\frac{f(x+h)-f(x)}{h}$$

$$\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}$$

I multiply it by its conjugate and I get:

$$\frac{x+h-x}{h(\sqrt[3]{x+h}+\sqrt[3]{x}}$$

I factor out terms and combine to get:

$$\frac{1}{2\sqrt[3]{x}}$$

Not at all the right answer, can somebody tell me where I've gone wrong?

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  • $\begingroup$ When you multiply by what you are calling the conjugate, the numerator isn't coming out right. $(\sqrt[3]{x + h} - \sqrt[3]{x})(\sqrt[3]{x + h} + \sqrt[3]{x}) \ne x + h - x$. $\endgroup$ – qaphla Jul 7 '14 at 19:08
  • $\begingroup$ $\sqrt[3]{x}^2 \neq x$ $\endgroup$ – Sten Jul 7 '14 at 19:08
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When you want to get '$x+h-x$' from '$\sqrt[3]{x+h}-\sqrt[3]{x}$' use formula:

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Not $\displaystyle a^3-b^3 \neq (a-b)(a+b)$.

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