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