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What is the derivative of #(6)/(x^3sqrtx)#?

Answer 1

Scarlett Allison

The answer is: #y'=-21/(x^4sqrtx)#.

The function can be written:

#y=6*x^(-3-1/2)rArry=6x^(-7/2)#

#y'=6*(-7/2)*x^(-7/2-1)=-21x^(-9/2)=-21/sqrt(x^9)=#

#=-21/(x^4sqrtx)#.

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Answer 2

Axel Alvarez

To find the derivative of ( \frac{6}{x^3\sqrt{x}} ), you can use the quotient rule. The quotient rule states that if you have a function ( \frac{u(x)}{v(x)} ), then its derivative is given by:

[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ]

Here, ( u(x) = 6 ) and ( v(x) = x^3\sqrt{x} ).

[ u'(x) = 0 ]

[ v'(x) = 3x^2\sqrt{x} + \frac{1}{2}x^3\cdot \frac{1}{\sqrt{x}} = 3x^2\sqrt{x} + \frac{x^3}{2\sqrt{x}} ]

Now, we apply the quotient rule:

[ \left(\frac{6}{x^3\sqrt{x}}\right)' = \frac{0\cdot x^3\sqrt{x} - 6\cdot \left(3x^2\sqrt{x} + \frac{x^3}{2\sqrt{x}}\right)}{(x^3\sqrt{x})^2} ]

[ = -\frac{6\cdot \left(3x^2\sqrt{x} + \frac{x^3}{2\sqrt{x}}\right)}{x^6} ]

[ = -\frac{18x^2\sqrt{x} + \frac{3x^3}{2\sqrt{x}}}{x^6} ]

[ = -\frac{18\sqrt{x}}{x^4} - \frac{3}{2x^{5/2}} ]

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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.