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

2 Answers
Jun 1, 2016

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

Explanation:

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}

Jun 1, 2016

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

Explanation:

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)