How do you differentiate #log_x3# =y ?

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Answer 1 · 2017-06-15T14:17:11

#dy/dx = -log_x3/(xlnx)#

You can approach this in two different ways:

(1) Changing the base of the logarithm, yields:

#log_x3 = ln3/lnx#

So:

#y=ln3/lnx#

#dy/dx = ln3 d/dx(1/lnx) = -ln3/(xln^2x)#

#dy/dx = -ln3/lnx 1/(xlnx) = - log_x3/(xlnx)#

(2) Based on the definition of logarithm:

#log_x3 = y <=> 3=x^y#

or:

#e^(ylnx) = 3#

Differentiating implicitly:

#d/dx(e^(ylnx)) = 0#

#e^(ylnx) d/dx(ylnx) = 0#

#e^(ylnx) (y/x +lnxdy/dx) = 0#

And since the exponential is never null:

#(y/x +lnxdy/dx) = 0#

#dy/dx = -y/(xlnx)#

Substituting #y# from the original equation:

#dy/dx = -log_x3/(xlnx)#