Question

Find the derivative of the following functions (it is to be understood that $a,b,c,d,p,q,r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) : $\frac{\sec x-1}{\sec x+1}$

Answer 1

Let $f\left(x\right)=\frac{\sec x-1}{\sec x+1}$
$=\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}=\frac{1-\cos x}{1+\cos x}$
Thus using quotient rule
$f^{\prime }\left(x\right)=\frac{(1+\cos x)\frac{d}{dx}(1-\cos x)-(1-\cos x)\frac{d}{dx}(1+\cos x)}{{(1+\cos x)}^2}$
$=\frac{(1+\cos x)\left(\sin x\right)-(1-\cos x)(-\sin x)}{{(1+\cos x)}^2}$
= $\frac{\sin x+\cos x\sin x+\sin x-\sin x\cos x}{{(1+\cos x)}^2}$
=$\frac{2\sin x}{{(1+\frac{1}{\sec x})}^2}=\frac{2\sin x}{\frac{{(\sec x+1)}^2}{{\sec }^{2}x}}$
= $\frac{2\sin x{\sec }^{2}x}{{(\sec x+1)}^2}=\frac{\frac{2\sin x}{\cos x}\sec x}{{(\sec x+1)}^2}=\frac{2\sec x\tan x}{{(\sec x+1)}^2}$