What is the first differential of #y=10^(3x-4)# ?

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Answer 1 · 2017-04-30T04:17:05

#dy/dx= 3ln10*10^(3x-4)#

#y=10^(3x-4)#

#lny = (3x-4)*ln10#

#1/y dy/dx = ln10*d/dx(3x-4)# [Implicit differentiation and standard differential]

#1/y dy/dx= ln10*(3-0)# [Power rule]

#dy/dx = 3ln10*y#

#= 3ln10*10^(3x-4)#

Answer 2 · 2017-04-30T15:27:50

A few additional observations

Notice that #10^(3x-4) = 10^(3x)/10^4#
So that #10^(3x-4) = 1000^(x)/10000#
Therefore if #y = 10^(3x-4)# then #y = 1000^(x)/10000#

This gives us #dy/dx = (1/10000)1000^x*ln(1000)#.
If we wish to simplify ln(1000), we have
#dy/dx = ((3ln10)/10000)1000^x#.

This answer is equal to the one given in the original response.