Question
Diffrentiate the following w.r.t.x
$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{4 x}-1}{e^{4 x}+1}\right) \\ & =\frac{\left(e^{4 x}+1\right) \cdot \frac{d}{d x}\left(e^{4 x}-1\right)-\left(e^{4 x}-1\right) \cdot \frac{d}{d x}\left(e^{4 x}+1\right)}{\left(e^{4 x}+1\right)^2} \\ & =\frac{\left(e^{4 x}+1\right)\left[e^{4 x} \cdot \frac{d}{d x}(4 x)-0\right]-\left(e^{4 x}-1\right)\left[e^{4 x} \cdot \frac{d}{d x}(4 x)+0\right]}{\left(e^{4 x}+1\right)^2}\end{aligned}$
$=\frac{\left(e^{4 x}+1\right) \cdot e^{4 x} \times 4-\left(e^{4 x}-1\right) \cdot e^{4 x} \times 4}{\left(e^{4 x}+1\right)^2 \text { } }$
$=\frac{4 e^{4 x}\left(e^{4 x}+1-e^{4 x}+1\right)}{\left(e^{4 x}+1\right)^2}=\frac{8 e^{4 x}}{\left(e^{4 x}+1\right)^2}$
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(i) the length of the perpendicular from the origin to the plane (ii) direction cosines of the normal.