Question

Find the derivative of
(i) 2x - $\frac{3}{4}$
(ii)(5x3+3x-1)(x-1)
(iii) x-3(5+3x) (iv)x5(3-6x-9)
(v) x-4(3-4x-5) (vi)$\frac{2}{x+1}-\frac{x^2}{3x-1}$

Answer 1

(i) Let f(x)=2x-$\frac{3}{4}$
f'(x)=$\frac{d}{dx}$ (2x-$\frac{3}{4}$)
=2$\frac{d}{dx}$(x) -$\frac{d}{dx}$($\frac{3}{4}$)
=2-0
=2

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(ii) Let f (x)=(5x3+3x-1) (x - 1)
By Leibnitz product rule,
ƒ'(x)=(5x3 +3x−1) = $\frac{d}{dx}$(x −1)+(x−1) = (5x3 +3x−1)
=(5x3+3x-1)(1)+(x-1)(5.3x2+3-0)
=(5x3+3x-1)+(x-1) (15x2+3)
=5x3+3x-1+15x3+3x-15x2-3
=20x3-15x2+6x-4

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(iii) Let f(x) = x-3 (5+3x)
By Leibnitz product rule,
ƒ′(x) = x-3$\frac{d}{dx}$(5+3x)+(5+3x)$\frac{d}{dx}$ (x-3)
= x-3 (0+3)+(5+3x)(−3x-3-1)
=x-3(3)+(5+3x)(-3x-4)
=3x-3-15x-4-9x-3
=-6x-3-15x-4
=${\frac{-3x^{-3}}{x}}$(2x+5)
=$-\frac{3}{x^4}$(5+2x)

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(iv) Let f (x) = x5(3-6x-9)
By Leibnitz product rule,
ƒ′(x) = x5-(3–6x-9)+(3–6x-9) $\frac{d}{dx}$(x5)
=x5 {0-6(-9)x-9-1}+(3–6x-9) (5x4)
= x5(54x-10)+15x4-30x-5
=54x-5+15x4-30x-5
= 24x-5+15x4
=$\frac{15x^4+24}{x^5}$

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(v) Let f (x) = x4(3-4x-5)
By Leibnitz product rule,
ƒ'(x) = x-4 $\frac{d}{dx}$(3-4x-5)+(3-4x-5)$\frac{d}{dx}$(x-4)
= x-4{0-4(-5) x-5-1}+(3-4x-5)(-4)x-4-1
x-4(20x-6)+(3-4x-5)(-4x-5)
=20x-10-12x-5+16x-10
=36x-10-12x-5
=$-\frac{12}{x^5}$ + $\frac{36}{x^{10}}$

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(vi) Let f (x) =$\frac{2}{x+1}-\frac{x^2}{3x-1}$
f '(x) =$\frac{d}{dx}$($\frac{2}{x+1}$) - $\frac{d}{dx}$($\frac{x^2}{3x-1}$)
By the quotient rule,
f'(x)=[(x+1)$\frac{d}{dx}$(2)-2$\frac{d}{dx}$(x+1) /(x+1)2]-[(3x-1)$\frac{d}{dx}$(x2)-x2$\frac{d}{dx}$(3x-1)]
=$[\frac{(x+1)(0)-2(1)}{(x+1)^2}]$-[$\frac{(3x-1)(2x)-(x^2)(3)}{(3x-1)^2}$]
=$-\frac{2}{(x+1)^2}$ - $[\frac{(3x-1)(2x)-(x^2)(3)}{(3x-1)^2}]$
=$-\frac{2}{(x+1)^2}$ -[$\frac{6x^2-2x-3x^2}{(3x-1)^2}$]
=$-\frac{2}{(x+1)^2}$ - [$\frac{3x^2-2x^2}{(3x-1)^2}$]
=$-\frac{2}{(x+1)^2}$ -x$\frac{(3x-2)}{(3x-1)^2}$