How do you differentiate # y=ln((6x-5)^6) # using the chain rule?

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Answer 1 · 2016-06-01T17:50:00

#frac{d}{dx}(ln ((6x-5)^6))=frac{36}{6x-5}#

#frac{d}{dx}(ln ((6x-5)^6))#

Applying chain rule,#frac{df(u)}{dx}=frac{df}{du}\cdot \frac{du}{dx}#

Let,#(6x-5)^6=u#
#=frac{d}{du}(ln (u))frac{d}{dx}((6x-5)^6)#

We know,
#frac{d}{du}(ln (u))=frac{1}{u}#
and,
#frac{d}{dx}((6x-5)^6)=36(6x-5)^5#

So,
#=\frac{1}{u}36\(6x-5\)^5#

substituting back,#\:u=\(6x-5\)^6#

#=\frac{1}{(6x-5)^6}36(6x-5)^5#

Simplifying it,we get,

#frac{36}{6x-5}#

Answer 2 · 2016-06-01T18:15:43

Use properties of logarithms to write #y=6ln(6x-5)# then us #d/dx(lnu) = 1/u (du)/dx#

#y = ln((6x-5)^6) = 6ln(6x-5)#

So,

#dy/dx = 6[1/(6x-5) d/dx(6x-5)]#

# = 6[1/(6x-5) (6)]#

# = 36/(6x-5)#