What is the derivative of $x^{\frac{3}{2}} {log}_{2} \sqrt{x + 1}$ $x^(3/2)log_2sqrt(x+1)$ ?

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What is the derivative of $x^{\frac{3}{2}} {log}_{2} \sqrt{x + 1}$ $x^(3/2)log_2sqrt(x+1)$ ?

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Answer 1 · 2018-03-13T13:02:04

$\frac{\sqrt{x}}{2 ln 2} [\frac{x}{x + 1} + 3 \cdot ln \sqrt{x + 1}]$ $(sqrtx)/(2ln2)[x/(x+1)+3*lnsqrt(x+1)]$

Explanation:

$N o t e : (1) {log}_{a} b = \frac{{log}_{e} b}{{log}_{e} a}$ $Note:color(red)((1)log_ab=(log_eb)/(log_ea))$
$(2) \frac{d}{d x} (u \cdot v) = u \cdot \frac{d v}{d x} + v \cdot \frac{d u}{d x}$ $color(blue)((2)d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx))$
$y = x^{\frac{3}{2}} {log}_{2} \sqrt{x + 1} = x^{\frac{3}{2}} [\frac{{log}_{e} \sqrt{x + 1}}{{log}_{e} 2}]$ $y=x^(3/2)log_2sqrt(x+1)=x^(3/2)[(log_esqrt(x+1))/log_e2]$
$y = \frac{1}{{log}_{e} 2} \cdot x^{\frac{3}{2}} \cdot {log}_{e} \sqrt{x + 1}$ $y=1/log_e2*x^(3/2)*log_esqrt(x+1)$
$y = \frac{1}{ln 2} \cdot x^{\frac{3}{2}} \cdot ln \sqrt{x + 1}$ $y=1/ln2*x^(3/2)*lnsqrt(x+1)$
$\frac{d y}{d x} = \frac{1}{ln 2} [x^{\frac{3}{2}} \frac{d}{d x} (ln \sqrt{x + 1}) + ln \sqrt{x + 1} \frac{d}{d x} (x^{\frac{3}{2}})]$ $(dy)/(dx)=1/ln2[x^(3/2)d/(dx)(lnsqrt(x+1))+lnsqrt(x+1)d/(dx)(x^(3/2))]$
$= \frac{1}{ln 2} [x^{\frac{3}{2}} \cdot \frac{1}{\sqrt{x + 1}} \cdot \frac{1}{2 \sqrt{x + 1}} + ln \sqrt{x + 1} \cdot \frac{3}{2} \cdot x^{\frac{1}{2}}]$ $=1/ln2[x^(3/2)*1/(sqrt(x+1))*1/(2sqrt(x+1))+lnsqrt(x+1)*3/2*x^(1/2)]$
$= \frac{1}{ln 2} [\frac{x \cdot \sqrt{x}}{2 (x + 1)} + \frac{3}{2} \cdot \sqrt{x} \cdot ln \sqrt{x + 1}]$ $=1/ln2[(x*sqrtx)/(2(x+1))+3/2*sqrtx*lnsqrt(x+1)]$
$= \frac{\sqrt{x}}{2 ln 2} [\frac{x}{x + 1} + 3 \cdot ln \sqrt{x + 1}]$ $=(sqrtx)/(2ln2)[x/(x+1)+3*lnsqrt(x+1)]$

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What is the derivative of $x^{\frac{3}{2}} {log}_{2} \sqrt{x + 1}$ $x^(3/2)log_2sqrt(x+1)$ ?

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