Question

Differentiate the function w.r.t.x:
$\rm \bigg(\frac{1}{x\sqrt{x}}-3x^{2} \bigg) \log x$

Answer 1

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Answer 2

Answer:

$\frac{d}{dx}(({x^{-\frac{3}{2}}-3x^2)log(x)) = x^{-\frac{5}{2}}-(\frac{3}{2x^{\frac{5}{2}}}-6x)log(x)-3x$

Step-by-step explanation:

By product rule, we have

$\frac{d}{dx}((x^{-\frac{3}{2}}-3x^2)log(x)) = \frac{d}{dx}(x^{-\frac{3}{2}}-3x^2)log(x) +(x^{-\frac{3}{2}}-3x^2)\frac{d}{dx}(log(x))$

Remember that $\frac{d}{dx}log(x) = \frac{1}{x}$ and, using the polynomial rule, we also have that

$\frac{d}{dx}(x^{-\frac{3}{2}}-3x^2) = \frac{-3}{2}x^{\frac{-5}{2}}-6x$

Now all we need to do is put it all together

$\frac{d}{dx}((x^{-\frac{3}{2}}-3x^2)log(x)) = \log(x)(\frac{-3}{2}x^{\frac{-5}{2}}-6x)+\frac{x^{\frac{-3}{2}}-3x^2}{x} = \\x^{\frac{-5}{2}}(\frac{-3}{2}\log(x)+1)+3x(2\log(x)-1)$

As wished.