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How do you simplify this problem? $$ \frac {\mathrm d}{\mathrm dx}\left[(3x+1)^3\sqrt{x}\right] $$ $$= \frac {(3x+1)^3}{2\sqrt {x}} + 9\sqrt{x} (3x+1)^2 $$ $$\frac{(3x+1)^2(21x+1)}{2\sqrt x} $$

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    that's as simple as it's gonna get. – 2017-02-07
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    =/ help me out please? – 2017-02-07
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    Well now it's too vague... what step don't you understand, the derivative or putting the terms over a common denominator? – 2017-02-07
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    How they simplify to get that answer – 2017-02-07
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    $\frac{a^3}{b}+a^2\cdot d=a^2(\frac{a}{b}+d)=(a^2)(\frac{a+b\cdot d}{b})$, take $a=(3x+1),b=2\sqrt x, d=9\sqrt x $ – 2017-02-07

3 Answers 3

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Multiplying top and bottom of the term $9\sqrt x(3x+1)^2$ by $2\sqrt x$ gives $$ \frac{18x(3x+1)^2}{2\sqrt x}.$$

Now can combine with the first term to get $$\frac{(3x+1)^3 + 18x(3x+1)^2}{2\sqrt x}.$$

Then we can factor a $(3x+1)^2$ out of the numerator, giving $$\frac{(3x+1)^2((3x+1)+18x)}{2\sqrt x} = \frac{(3x+1)^2(21x+1)}{2\sqrt x} $$

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    Thanks a lot for your help! – 2017-02-07
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Use the product rule of differentiation, that is $\mathrm {d}(uv) = u\mathrm {dv} + v\mathrm {du} $. We thus get, $$\frac {d}{dx}[(3x+1)^3\sqrt{x}] $$ $$= (3x+1)^3\frac {d}{dx}[\sqrt {x}] + \sqrt {x}\frac {d}{dx}[(3x+1)^3] $$ $$= \frac {(3x+1)^3}{2\sqrt {x}} + 3 (3x+1)^2 (3)\sqrt{x} $$ $$= (3x+1)^2 [\frac {3x+1}{2\sqrt {x}} + 9\sqrt {x}] $$ $$= (3x+1)^2 [\frac {(3x+1) + 2\sqrt {x}[9\sqrt {x}]}{2\sqrt {x}}] $$ $$= \frac{(3x+1)^2 (21x+1)}{2\sqrt {x}} $$ Hope it helps.

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Just offering another tool here, logarithmic differentiation. A specific use of implicit differentiation.

Put $y= (3x+1)^3\sqrt x$

$$\log y = 3\log(3x+1) + \frac 12 \log x$$

Observe that $\frac{d}{dx} \log y = \frac{d}{dy} (\log y) \cdot \frac{dy}{dx}$ by the chain rule. Use $y'$ to represent $\frac{dy}{dx}$ for clarity.

$$\frac{y'}{y} = \frac{(3)(3)}{3x+1} + \frac{1}{2x}$$

$$y' = (3x+1)^3\sqrt x (\frac{9}{3x+1} + \frac{1}{2x})$$

(I'll leave any further simplification to you).