Differentiate w.r.t x:

Question:

Differentiate w.r.t x:

$\sqrt{\frac{1+\sin x}{1-\sin x}}$

 

Solution:

Let $y=\sqrt{\frac{1+\sin x}{1-\sin x}}, u=1+\sin x, v=1-\sin x, z=\frac{1+\sin x}{1-\sin x}$

Formula :

$\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$

According to the quotient rule of differentiation

If $\mathrm{z}=\frac{\mathrm{u}}{\mathrm{v}}$

$\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

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

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

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

According to the chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dx}}$

$=\left[\frac{1}{2} \times\left(\frac{1+\sin x}{1-\sin x}\right)^{\frac{1}{2}-1}\right] \times\left[\frac{2 \cos x}{(1-\sin x)^{2}}\right]$

$=\left[\frac{\cos x}{1} \times\left(\frac{1+\sin x}{1}\right)^{-\frac{1}{2}}\right] \times\left[\frac{1}{(1-\sin x)^{2-\frac{1}{2}}}\right]$

$=\left[\cos x \times(1+\sin x)^{-\frac{1}{2}}\right] \times(1-\sin x)^{-\frac{3}{2}}$

 

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