How do you differentiate #y=ln(x^2+1)-log_2(5x)#?

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Answer 1 · 2017-08-16T03:49:22

Please see below.

#d/dx(ln(x^2+1))# use #d/dx(lnu) = 1/u (du)/dx# so

#d/dx(ln(x^2+1)) = (2x)/(x^2+1)#

#d/dx(log_2(5x)) = d/dx(log_2(5))+d/dx(log_2(x))#

# = 0 + d/dx(log_2(x))#

# = 1/(xln2)#

So, for #y = ln(x^2+1)-log_2(5x)#, we have

#dy/dx = (2x)/(x^2+1)-1/(xln2)#

Note

#log_2(x) = lnx/ln2 = 1/ln2 * lnx#

So the derivtaive is #1/ln2 * 1/x = 1/(xln2)#