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I'm trying to differentiate the following function:

$f(x)=4x^2\sqrt{x}+6\sqrt{x}+\frac{8}{\sqrt{x}}$

I know the result is $f'(x)=10x\sqrt{x}+\frac{3}{\sqrt{x}}-\frac{4}{x\sqrt{x}}$

However, I have no idea about how to do it. Please, can anyone explain me what are the steps here?

superjim
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  • Notice $4x^2\sqrt{x} = 4x^2x^{1/2} = 4x^{5/2}$. You can use the power rule when differentiating. For the last term, it can be written as $8x^{-1/2}$. – MathNewbie Nov 22 '15 at 15:40

2 Answers2

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One may recall that $$ (x^\alpha)'=\alpha x^{\alpha-1} \tag1 $$ Then rewrite $f(x)$ as $$ f(x)=4x^{5/2}+6x^{1/2}+8x^{-1/2} \tag2 $$ and apply $(1)$ to $(2)$.

Olivier Oloa
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  • Thanks! But wouldn't it be $10x^\frac{3}{2}+3-4$? Why do I have $\frac{3}{\sqrt{x}}$ and $\frac{4}{x\sqrt{x}}$ in the results? – superjim Nov 22 '15 at 15:54
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Hint use the product rule . Which states $(uv)'=u'v+v'u$ consider $u=4x^2,v=\sqrt{x}$ and same for the next two.

Archis Welankar
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