Question

If $y=\frac{a^{{\cos }^{-1}x}}{1+a^{{\cos }^{-1}x}}$ and $z=a^{{\cos }^{-1}x}$, then $\frac{dy}{dz}$ is equal to

• A. $\frac{1}{1+a^{{\cos }^{-1}x}}$
• B. $-\frac{1}{1+a^{{\cos }^{-1}x}}$
• C. $\frac{1}{{(1+a^{{\cos }^{-1}x})}^2}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{{(1+a^{{\cos }^{-1}x})}^2}$

$y=\frac{a^{{\cos }^{-1}x}}{1+a^{\cos (-1x)}},z=a^{{\cos }^{-1}x}$

$y=\frac{z}{1+z}$

$\therefore \frac{dy}{dz}=\frac{(1+z)\left(1\right)-\left(z\right)\left(1\right)}{{(1+z)}^2}$

$=\frac{1+z-z}{{(1+z)}^2}=\frac{1}{{(1+z)}^2}$

$=\frac{1}{{(1+a^{{\cos }^{-1}x})}^2}$