Differentiate w.r.t x:

Question:

Differentiate w.r.t x:

$\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$

 

Solution:

Let $y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}, u=e^{2 x}+e^{-2 x}, v=e^{2 x}-e^{-2 x}$

Formula :

$\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$

According to the quotient rule of differentiation

If $y=\frac{u}{v}$

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

$=\frac{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right) \times\left(2 \mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{-2 \mathrm{x}}\right)-\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}\right) \times\left(2 \mathrm{e}^{2 \mathrm{x}}+2 \mathrm{e}^{-2 \mathrm{x}}\right)}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

$=\frac{2\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}-2\left(\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

$=\frac{2\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}+\mathrm{e}^{2 \mathrm{x}}+\mathrm{e}^{-2 \mathrm{x}}\right)\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}-\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

$\left(a^{2}-b^{2}=(a-b)(a+b)\right.$

$=\frac{2\left(2 \mathrm{e}^{2 \mathrm{x}}\right)\left(-2 \mathrm{e}^{-2 \mathrm{x}}\right)}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

$=\frac{-8}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

 

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