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How do you find the derivative of #y=3x^4 - sqrt(x)/9 + 4/x^3 + 6xsqrt(x)#?

Answer 1

Charles Anctil

Refer to explanation

We rewrite it as follows

#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)#

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Answer 2

Christopher Agresta

Refer Explanation section

As far as possible I simplified the process

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Answer 3

Adam Adkins

To find the derivative of the given function, ( y = 3x^4 - \frac{\sqrt{x}}{9} + \frac{4}{x^3} + 6x\sqrt{x} ), you differentiate each term individually using the rules of differentiation.

( \frac{d}{dx} [3x^4] = 12x^3 )

( \frac{d}{dx} [\frac{\sqrt{x}}{9}] = \frac{1}{9} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{18x^{1/2}} )

( \frac{d}{dx} [\frac{4}{x^3}] = -12x^{-4} )

( \frac{d}{dx} [6x\sqrt{x}] = 6\sqrt{x} + 6x \cdot \frac{1}{2\sqrt{x}} = 6\sqrt{x} + \frac{3}{\sqrt{x}} )

Therefore, the derivative of the function is:

( y' = 12x^3 - \frac{1}{18x^{1/2}} - 12x^{-4} + 6\sqrt{x} + \frac{3}{\sqrt{x}} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.