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Question: If f and g be two functions such that f(x) ¹ 0, g(x) ¹ f ¢(x) ¹ 0 and g ¢(x) ¹ 0 for all x, then (1...

If f and g be two functions such that f(x) ¹ 0, g(x) ¹ f ¢(x) ¹ 0 and g ¢(x) ¹ 0 for all x, then

(1) (fg)′=f′g′\left( \frac{f}{g} \right)^{'} = \frac{f'}{g'} (2) (fg)fg′=f′f+g′g\frac{(fg)}{fg}^{'} = \frac{f'}{f} + \frac{g'}{g}

(3) (f+g)f+g′=f′f+g′g\frac{(f + g)}{f + g}^{'} = \frac{f'}{f} + \frac{g'}{g} (4) (f/g)f/g′=f′f−g′g\frac{(f/g)}{f/g}^{'} = \frac{f'}{f} - \frac{g'}{g}

Which of these statement are correct

A

1 & 2

B

2 & 3

C

2 & 4

D

3 & 4

Answer

2 & 4

Explanation

Solution

(1) (fg)′=gf′−fg′g2≠fg\left( \frac{f}{g} \right)^{'} = \frac{gf' - fg'}{g^{2}} \neq \frac{f}{g}

(2) (fg)fg′=fg′+gf′fg=g′g+f′f\frac{(fg)}{fg}^{'} = \frac{fg' + gf'}{fg} = \frac{g'}{g} + \frac{f'}{f} (correct)

(3) (f+g)′f+g=f′+g′f+g\frac{(f + g)'}{f + g} = \frac{f' + g'}{f + g}

(4)(f/g)′f/g=(gf′−fg′)/g2f/g=f′f−g′g\frac{(f/g)'}{f/g} = \frac{(gf' - fg')/g^{2}}{f/g} = \frac{f'}{f} - \frac{g'}{g} (correct)