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How do you find the derivative using the Quotient rule for #f(z)= (z^2+1)/(sqrtz)#?

Answer 1

Paisley Johnson

#y'=(3z^2-1)/(2zsqrtz)#.

In this way:

#y'=(2z*sqrtz-(z^2+1)*1/(2sqrtz))/(sqrtz)^2=#

#=((2z*sqrtz*2sqrtz-z^2-1)/(2sqrtz))/z=(4z^2-z^2-1)/(2zsqrtz)=#

#=(3z^2-1)/(2zsqrtz)#.

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Answer 2

Axel Alvarez

To find the derivative of ( f(z) = \frac{{z^2 + 1}}{{\sqrt{z}}} ) using the Quotient Rule, you apply the formula:

[ f'(z) = \frac{{g(z) \cdot f'(z) - f(z) \cdot g'(z)}}{{[g(z)]^2}} ]

Where ( f(z) = z^2 + 1 ) and ( g(z) = \sqrt{z} ).

First, find the derivatives of ( f(z) ) and ( g(z) ):

[ f'(z) = 2z ] [ g'(z) = \frac{1}{2\sqrt{z}} ]

Now, plug these into the Quotient Rule formula:

[ f'(z) = \frac{{(\sqrt{z})(2z) - (z^2 + 1)(\frac{1}{2\sqrt{z}})}}{{(\sqrt{z})^2}} ]

Simplify:

[ f'(z) = \frac{{2z\sqrt{z} - \frac{z^2 + 1}{2\sqrt{z}}}}{{z}} ]

[ f'(z) = \frac{{2z\sqrt{z} - \frac{z^2}{2\sqrt{z}} - \frac{1}{2\sqrt{z}}}}{{z}} ]

[ f'(z) = \frac{{4z^{\frac{3}{2}} - z - \frac{1}{2z^{\frac{1}{2}}}}}{{2z}} ]

[ f'(z) = 2z^{\frac{1}{2}} - \frac{1}{2z^{\frac{1}{2}}} - \frac{1}{2z} ]

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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.