Difference between revisions of "Derivative of arcsinh"
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==Proof== | ==Proof== | ||
| + | From the definition, | ||
| + | $$\mathrm{arcsinh}(z)=\log \left(z + \sqrt{1+z^2} \right).$$ | ||
| + | Using [[derivative of logarithm]] and the [[chain rule]], | ||
| + | $$\begin{array}{ll} | ||
| + | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z} \log \left(z + \sqrt{1+z^2} \right) \\ | ||
| + | &= \dfrac{1}{z+\sqrt{1+z^2}} \dfrac{\mathrm{d}}{\mathrm{d}z} \Bigg[ z + \sqrt{1+z^2} \Bigg] \\ | ||
| + | &= \dfrac{1+\frac{z}{\sqrt{1+z^2}}}{z+\sqrt{1+z^2}} \\ | ||
| + | &= \dfrac{\sqrt{1+z^2}+z}{z\sqrt{1+z^2}+1+z^2} \\ | ||
| + | &= \dfrac{\sqrt{1+z^2}+z}{z\sqrt{1+z^2}+1+z^2} \left( \dfrac{\sqrt{1+z^2}-z}{\sqrt{1+z^2}-z} \right) \\ | ||
| + | &= \dfrac{1}{z(1+z^2)+\sqrt{1+z^2}+z^2\sqrt{1+z^2}-z^2\sqrt{1+z^2}-z-z^3} \\ | ||
| + | &= \dfrac{1}{\sqrt{1+z^2}}, | ||
| + | \end{array}$$ | ||
| + | as was to be shown. | ||
==References== | ==References== | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proven]] |
Latest revision as of 00:08, 16 September 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) = \dfrac{1}{\sqrt{1+z^2}},$$ where $\mathrm{arcsinh}$ denotes the inverse hyperbolic sine.
Proof
From the definition, $$\mathrm{arcsinh}(z)=\log \left(z + \sqrt{1+z^2} \right).$$ Using derivative of logarithm and the chain rule, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z} \log \left(z + \sqrt{1+z^2} \right) \\ &= \dfrac{1}{z+\sqrt{1+z^2}} \dfrac{\mathrm{d}}{\mathrm{d}z} \Bigg[ z + \sqrt{1+z^2} \Bigg] \\ &= \dfrac{1+\frac{z}{\sqrt{1+z^2}}}{z+\sqrt{1+z^2}} \\ &= \dfrac{\sqrt{1+z^2}+z}{z\sqrt{1+z^2}+1+z^2} \\ &= \dfrac{\sqrt{1+z^2}+z}{z\sqrt{1+z^2}+1+z^2} \left( \dfrac{\sqrt{1+z^2}-z}{\sqrt{1+z^2}-z} \right) \\ &= \dfrac{1}{z(1+z^2)+\sqrt{1+z^2}+z^2\sqrt{1+z^2}-z^2\sqrt{1+z^2}-z-z^3} \\ &= \dfrac{1}{\sqrt{1+z^2}}, \end{array}$$ as was to be shown.
