Express as a natural log,then differentiate with respect to x,Find dy/dx (a)y=(2x-3)^5/(x²+1)^6√(x^3-2x). (B)y=log4 2x (c) y=x^(2x)?

Question

962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-18T02:05:44

See the explanation section.

Here is my best guess what the intended functions are:

(a)#y=(2x-3)^5/(x²+1)^6sqrt(x^3-2x)#.

#ln y=ln ((2x-3)^5/(x²+1)^6sqrt(x^3-2x))#

# = 5ln(2x-3)-6ln(x^2+1) + 1/2ln(x^3-2x)#

#1/y dy/dx = 5[1/(2x-3) * 2] -6[1/(x^2+1) * 2x] + 1/2 [1/(x^3-2x) * (3x^2-2)]#

# = [10/(2x-3)-(12x)/(x^2+1)+(3x^2-2)/(2(x^3-2x))]#

#dy/dx = y[10/(2x-3)-(12x)/(x^2+1)+(3x^2-2)/(2(x^3-2x))]#

# = (2x-3)^5/(x²+1)^6sqrt(x^3-2x)[10/(2x-3)-(12x)/(x^2+1)+(3x^2-2)/(2(x^3-2x))]#

(B)#y=log_4 2x#

#y = ln(2x)/ln4 = 1/ln4[ln2+lnx]#

#dy/dx = 1/ln4 [0+1/x] = 1/(xln4)#

(c) #y=x^(2x)#

#lny = lnx^(2x)= 2xlnx#

#1/y dy/dx = 2lnx+2x(1/x) = 2lnx+2#

#dy/dx = y[2lnx+2]#

# = x^(2x) [2lnx+2]#