Giải f(r_{3})=frac{tanh(sigma(r_{s}+R))-tanh(sigma(r_{s}-R))}{2tanh(sigmaR)}

$$f ( r _ { 3 } ) = \frac { \tanh ( \sigma ( r _ { s } + R ) ) - \tanh ( \sigma ( r _ { s } - R ) ) } { 2 \tanh ( \sigma R ) }$$

Tính giá trị

$$\frac{\tanh(\sigma \left(r_{s}+R\right))-\tanh(\sigma \left(r_{s}-R\right))}{2\tanh(R\sigma )}$$

Lấy vi phân theo σ

$$\frac{R\tanh(R\sigma )\left(\operatorname{sech}(\sigma \left(r_{s}+R\right))\right)^{2}+r_{s}\tanh(R\sigma )\left(\operatorname{sech}(\sigma \left(r_{s}+R\right))\right)^{2}-R\tanh(\sigma \left(r_{s}+R\right))\left(\operatorname{sech}(R\sigma )\right)^{2}+R\tanh(R\sigma )\left(\operatorname{sech}(\sigma \left(r_{s}-R\right))\right)^{2}+R\tanh(\sigma \left(r_{s}-R\right))\left(\operatorname{sech}(R\sigma )\right)^{2}-r_{s}\tanh(R\sigma )\left(\operatorname{sech}(\sigma \left(r_{s}-R\right))\right)^{2}}{2\left(\tanh(R\sigma )\right)^{2}}$$