There are certain derivatives you'll just have to memorize. For these problems, we have to know:

#d/(dx)(log_10x) = d/(dx)(lnx/(ln10)) = 1/(x*ln10)#

(could be derived from the change of base law!)

#d/(dx)a^(bx) = d/(dx)(a^x)^b = b*(a^x)^(b-1)*d/(dx)(a^x)#

#d/(dx)(a^x) = a^xlna#

(must memorize this one)

#d/(dx)a^(bx) = b*(a^x)^(b-1)*a^xlna = ba^(bx-x)*a^xlna#

# = ba^(bx)lna#

(can be derived)

Then there are all those inverse trig derivatives. Geez.

#d/(dx)(arc##sinx")" = 1/(sqrt(1-x^2)#

#d/(dx)(arc##cosx")" = -1/(sqrt(1-x^2)#

#d/(dx)(arc##tanx")" = 1/(1+x^2)#

#d/(dx)(arc##cotx")" = -1/(1+x^2)#

#d/(dx)(arc##secx")" = 1/(|x|sqrt(x^2-1))#

#d/(dx)(arc##cscx")" = -1/(|x|sqrt(x^2-1))#

Notice how #cosx#, #cotx#, and #cscx# are all negative, and otherwise the other trig function (#sinx#, #tanx#, #secx#) is the same.

That's all you need to do these.

**a)**

#d/(dx)(log_4(6x-4)) = d/(dx)[ln(6x-4)]/[ln4] = 6/((6x-4)ln4)#

**b)**

You will need the quotient rule in here somewhere, along with the chain rule.

#d/(dx)[log_3(x^2/(x-4))] = d/(dx)[[ln(x^2/(x-4))]/[ln3]]#

#= 1/(x^2/cancel(x-4))*[((x-4)(2x) - x^2(1))/(x-4)^cancel(2)]*1/ln3#

Multiply the numerator out, subtract the #x^2# terms, and factor out #x#:

#= 1/x^cancel(2)*[(cancel(x)(x-8))/(x-4)]*1/ln3#

#= 1/x*[(x-8)/(x-4)]*1/ln3#

#= (x-8)/(x-4)*1/(xln3)#

**c)**

You can just use the product rule here, with what we've derived above. There are also some creative exponent rules here.

#d/(dx)(x*4^(-2x)) = x*(-2*4^(-2x)ln4) + 4^(-2x)*1#

#= 4^(-2x)[(x*(-2ln4) + 1)]#

#= 4^(-2x)(1 - 2xln4)#

or

#16^(-x)(1 - 2xln4)#

**d)**

Reference the above, and use the chain rule.

#d/(dx)(arccos(4sqrtx))#

#= -1/(sqrt(1-16x))*4*1/(2sqrtx)#

#= -cancel(4)^(2)/(cancel(2)sqrtxsqrt(1-16x))#

#= -2/(sqrtxsqrt(1-16x))#

**e)**

Again, reference the above and use the chain rule.

#d/(dx)[xarctan(4x-6)] = x*[1/(1+(4x-6)^2)]*4 + arctan(4x-6)*1#

#= (4x)/(1+(4x-6)^2) + arctan(4x-6)#

or from Wolfram Alpha:

#= (4x)/((6-4x)^2 + 1) - arctan(6-4x)#

because #arctan(u) = -arctan(-u)#.