Question
Diffrentiate the following w.r.t.x
$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$
$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$
Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\right) \\ & =\frac{\left(e^{\sqrt{x}}-1\right) \frac{d}{d x}\left(e^{\sqrt{x}}+1\right)-\left(e^{\sqrt{x}}+1\right) \frac{d}{d x}\left(e^{\sqrt{x}}-1\right)}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\left(e^{\sqrt{x}}-1\right)\left[e^{\sqrt{x}} \cdot \frac{d}{d x}(\sqrt{x})+0\right]-\left(e^{\sqrt{x}}+1\right)\left[e^{\sqrt{x}} \cdot \frac{d}{d x}(\sqrt{x})-0\right]}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\left(e^{\sqrt{x}}-1\right)\left[e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}\right]-\left(e^{\sqrt{x}}+1\right)\left[e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}\right]}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\frac{e^{\sqrt{x}}}{2 \sqrt{x}}\left(e^{\sqrt{x}}-1-e^{\sqrt{x}}-1\right)}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{-e^{\sqrt{x}}}{\sqrt{x}\left(e^{\sqrt{x}}-1\right)^2} .\end{aligned}$
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$\int_0^1 t^5 \sqrt{1-t^2} d t$