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How do you find the instantaneous rate of change of #w# with respect to #z# for #w=1/z+z/2#?

Answer 1

Scarlett Allison

#(dw)/dz = -1/z^2 + 1/2#

Justification

#(dw)/dz = d/dz (1/z + z/2)#

first configuration.

#(dw)/dz = d / dz ( 1/z) + d /dz ( z/2)#

A sum's derivative is equal to the total of its derivatives.

#(dw)/dz = d/dz (z^-1) + 1/2 d/dz (z)#

First part: A function #f(z) = c/(z^n)# with #c# constant can also be written as #f(z) = cz^(-n)# Second part: #d/dz cf(z) = c d/dz f(z)# if c is constant.

#(dw)/dz = -1*z^ -2 + 1/2*1#

Use of the power rule: #d/dz z^n = nz^(n-1)#. Then #d/dz z = d/dz z^1 = z^0 = 1#

#(dw)/dz = -z^ -2 + 1/2#

Identity multiplicative postulate.

#(dw)/dz = -1/z^2 + 1/2#

A function written as #f(z) = cz^(-n)# can also be written #f(z) = c/(z^n)#

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Answer 2

Evelyn Agard

To find the instantaneous rate of change of ( w ) with respect to ( z ) for ( w = \frac{1}{z} + \frac{z}{2} ), you would differentiate ( w ) with respect to ( z ) using the rules of differentiation. This yields ( \frac{dw}{dz} = -\frac{1}{z^2} + \frac{1}{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.