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

1 Answer

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

Explanation :

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

Initial set-up.

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

The derivative of a sum is equal to the sum of the 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#

Multiplicative identity 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)#