Find the derivative of
Question:

Find the derivative of

(i) $2 x-\frac{3}{4}$

(ii) $\left(5 x^{3}+3 x-1\right)(x-1)$

(iii) $x^{-3}(5+3 x)$

(iv) $x^{5}\left(3-6 x^{-9}\right)$

(v) $x^{-4}\left(3-4 x^{-5}\right)$

(vi) $\frac{2}{x+1}-\frac{x^{2}}{3 x-1}$

Solution:

(i) Let $f(x)=2 x-\frac{3}{4}$

$f^{\prime}(x)=\frac{d}{d x}\left(2 x-\frac{3}{4}\right)$

$=2 \frac{d}{d x}(x)-\frac{d}{d x}\left(\frac{3}{4}\right)$

$=2-0$

$=2$

(ii) Let $f(x)=\left(5 x^{3}+3 x-1\right)(x-1)$

By Leibnitz product rule,

$f^{\prime}(x)=\left(5 x^{3}+3 x-1\right) \frac{d}{d x}(x-1)+(x-1) \frac{d}{d x}\left(5 x^{3}+3 x-1\right)$

$=\left(5 x^{3}+3 x-1\right)(1)+(x-1)\left(5.3 x^{2}+3-0\right)$

$=\left(5 x^{3}+3 x-1\right)+(x-1)\left(15 x^{2}+3\right)$

$=5 x^{3}+3 x-1+15 x^{3}+3 x-15 x^{2}-3$

$=20 x^{3}-15 x^{2}+6 x-4$

(iii) Let $f(x)=x^{-3}(5+3 x)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{-3} \frac{d}{d x}(5+3 x)+(5+3 x) \frac{d}{d x}\left(x^{-3}\right)$

$=x^{-3}(0+3)+(5+3 x)\left(-3 x^{-3-1}\right)$

$=x^{-3}(3)+(5+3 x)\left(-3 x^{-4}\right)$

$=3 x^{-3}-15 x^{-4}-9 x^{-3}$

$=-6 x^{-3}-15 x^{-4}$

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

$=\frac{-3 x^{-3}}{x}(2 x+5)$

$=\frac{-3}{x^{4}}(5+2 x)$

(iv) Let $f(x)=x^{5}\left(3-6 x^{-9}\right)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{5} \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x}\left(x^{5}\right)$

$=x^{5}\left\{0-6(-9) x^{-9-1}\right\}+\left(3-6 x^{-9}\right)\left(5 x^{4}\right)$

$=x^{5}\left(54 x^{-10}\right)+15 x^{4}-30 x^{-5}$

$=54 x^{-5}+15 x^{4}-30 x^{-5}$

$=24 x^{-5}+15 x^{4}$

$=15 x^{4}+\frac{24}{x^{5}}$

(v) Let $f(x)=x^{-4}\left(3-4 x^{-5}\right)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{-4} \frac{d}{d x}\left(3-4 x^{-5}\right)+\left(3-4 x^{-5}\right) \frac{d}{d x}\left(x^{-4}\right)$

$=x^{-4}\left\{0-4(-5) x^{-5-1}\right\}+\left(3-4 x^{-5}\right)(-4) x^{-4-1}$

$=x^{-4}\left(20 x^{-6}\right)+\left(3-4 x^{-5}\right)\left(-4 x^{-5}\right)$

$=20 x^{-10}-12 x^{-5}+16 x^{-10}$

$=36 x^{-10}-12 x^{-5}$

$=-\frac{12}{x^{5}}+\frac{36}{x^{10}}$

(vi) Let $f(x)=\frac{2}{x+1}-\frac{x^{2}}{3 x-1}$

$f^{\prime}(x)=\frac{d}{d x}\left(\frac{2}{x+1}\right)-\frac{d}{d x}\left(\frac{x^{2}}{3 x-1}\right)$

By quotient rule,

$f^{\prime}(x)=\left[\frac{(x+1) \frac{d}{d x}(2)-2 \frac{d}{d x}(x+1)}{(x+1)^{2}}\right]-\left[\frac{(3 x-1) \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(3 x-1)}{(3 x-1)^{2}}\right]$

$=\left[\frac{(x+1)(0)-2(1)}{(x+1)^{2}}\right]-\left[\frac{(3 x-1)(2 x)-\left(x^{2}\right)(3)}{(3 x-1)^{2}}\right]$

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

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

$=\frac{-2}{(x+1)^{2}}-\frac{x(3 x-2)}{(3 x-1)^{2}}$

 

 

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