# 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)?

Sep 18, 2015

See the explanation section.

#### Explanation:

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))

$= 5 \ln \left(2 x - 3\right) - 6 \ln \left({x}^{2} + 1\right) + \frac{1}{2} \ln \left({x}^{3} - 2 x\right)$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 5 \left[\frac{1}{2 x - 3} \cdot 2\right] - 6 \left[\frac{1}{{x}^{2} + 1} \cdot 2 x\right] + \frac{1}{2} \left[\frac{1}{{x}^{3} - 2 x} \cdot \left(3 {x}^{2} - 2\right)\right]$

$= \left[\frac{10}{2 x - 3} - \frac{12 x}{{x}^{2} + 1} + \frac{3 {x}^{2} - 2}{2 \left({x}^{3} - 2 x\right)}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left[\frac{10}{2 x - 3} - \frac{12 x}{{x}^{2} + 1} + \frac{3 {x}^{2} - 2}{2 \left({x}^{3} - 2 x\right)}\right]$

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

(B)$y = {\log}_{4} 2 x$

$y = \ln \frac{2 x}{\ln} 4 = \frac{1}{\ln} 4 \left[\ln 2 + \ln x\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} 4 \left[0 + \frac{1}{x}\right] = \frac{1}{x \ln 4}$

(c) $y = {x}^{2 x}$

$\ln y = \ln {x}^{2 x} = 2 x \ln x$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \ln x + 2 x \left(\frac{1}{x}\right) = 2 \ln x + 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left[2 \ln x + 2\right]$

$= {x}^{2 x} \left[2 \ln x + 2\right]$