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)?

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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) $y=x^(2x)$

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

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

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

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