How do you find the derivative of $y = 3 x^{4} - \frac{\sqrt{x}}{9} + \frac{4}{x^{3}} + 6 x \sqrt{x}$ $y=3x^4 - sqrt(x)/9 + 4/x^3 + 6xsqrt(x)$ ?

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Answer 1 · 2015-09-15T02:00:16

Refer to explanation

We rewrite it as follows

$y = 3 x^{4} - \frac{1}{9} \cdot x^{\frac{1}{2}} + 4 \cdot x^{- 3} + 6 x^{\frac{3}{2}}$ $y=3x^4-1/9*x^(1/2)+4*x^-3+6x^(3/2)$

hence the derivative is

$y'=12x^3-1/9*(1/2)x^(-1/2)-12x^-4+6*3/2x^(1/2)=> y'=12x^3-1/18x^(-1/2)-12x^-4-9x^(1/2)$

Answer 2 · 2015-09-15T02:19:10

Refer Explanation section

As far as possible I simplified the process

Follow the steps

How do you find the derivative of y=3x^4 - sqrt(x)/9 + 4/x^3 + 6xsqrt(x)y=3x4−√x9+4x3+6x√xy=3x^4 - sqrt(x)/9 + 4/x^3 + 6xsqrt(x)?