Question

Find the derivative of:

(i) 2x - $\frac{3}{4}$
(ii) $(5x^3+3x-1)$ (x-1)

(iii) $x^{-3}$ (5+3x)
(iv) $x^5(3-6x^{-9})$

(v) $x^{-4}(3-4x^{-5})$
(vi) $\frac{2}{x+1}-\frac{x^2}{3x-1}$

Answer 1

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

$\therefore$ f'(x) = $\frac{d}{dx}(2x-\frac{3}{4})$

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

= 2 $\times$ 1 - 0 = 2.

(ii) Here f(x) = $(5x^3+3x-1)$ (x-1)

$\therefore$ f'(x) = $\frac{d}{dx}\left[\right(5x^3+3x-1\left)\right(x-1\left)\right]$

= $(5x^3+3x-1)\frac{d}{dx}(x-1)+(x-1)\frac{d}{dx}(5x^3+3x-1)$

= $(5x^3+3x-1)\times 1+(x-1)(15x^2+3)$

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

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

(iii) Here $f\left(x\right)=x^{-3}(5+3x)$

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

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

= $\frac{3}{x^3}[1-\frac{5+3x}{x}]=\frac{3}{x^3}\left[\frac{x-5-3x}{x}\right]$

= $\frac{-3}{x_4}$(5+2x).

(iv) Here f(x) = $x^5(3-6x^{-9})$

$\therefore$ f'(x) = $\frac{d}{dx}\left[x^5\right(3-6x^{-9}\left)\right]$

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

= $x^5\left(54x^{-10}\right)+(3-6x^{-9})\times 5x^4$

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

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

(v) Here f(x) =$x^{-4}(3-4x^{-5})$

$\therefore$ f'(x) = $\frac{d}{dx}\left[x^{-4}\right(3-4x^{-5}\left)\right]$

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

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

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

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

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

$\therefore$ f'(x) = $\frac{d}{dx}[\frac{2}{x+1}-\frac{x^2}{3x-1}]$

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

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

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

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