Question

If $f\left(x\right)={\left(\frac{x^a}{x^b}\right)}^{a+b}.{\left(\frac{x^b}{x^c}\right)}^{b+c}.{\left(\frac{x^c}{x^a}\right)}^{c+a}$ then $f^{\prime }\left(x\right)$ is equal to

• A. $1$
• B. $0$
• C. $x^{a+b+c}$
• D. None of these

Answer 1

The correct option is B $0$

$f\left(x\right)={\left(\frac{x^a}{x^b}\right)}^{(a+b)}.{\left(\frac{x^b}{x^c}\right)}^{(b+c)}.{\left(\frac{x^c}{x^a}\right)}^{(c+a)}$

$={\left(x^{(a-b)}\right)}^{(a+b)}{\left(x^{(b-c)}\right)}^{(b+c)}.{\left(x^{(c-a)}\right)}^{(c+a)}=x^{(a^2-b^2)}.x^{(b^2-c^2)}.x^{(c^2-a^2)}=x^{(a^2-b^2+b^2-c^2+c^2-a^2)}=x^0=1\therefore f^{\prime }\left(x\right)=0$