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Question: \[\frac{(fg)'}{fg} = \frac{fg' + gf'}{fg} = \frac{g'}{g} + \frac{f'}{f}\] \(\frac{(f/g)'}{f/g}\)=\(...

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

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

A

f '(1/2) = f '(–1/2)

B

f '(x) = f '(–1/2)

C

f (–1/2) = f(1/2)

D

f (1/2) = f '(–1/2)

Answer

f (–1/2) = f(1/2)

Explanation

Solution

f(x) = 4x8 Ž f ¢(x) = 32x7

Here f(1/2) = f(–1/2) = 4(1/2)8