Difference between revisions of "Derivative of arcsin"

From specialfunctionswiki
Jump to: navigation, search
(Proof)
 
(4 intermediate revisions by the same user not shown)
Line 5: Line 5:
  
 
==Proof==
 
==Proof==
If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ to get
+
If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use [[implicit differentiation]] with respect to $z$ and the [[derivative of sine]] to get
$$\cos(\theta)\theta'=1.$$
+
$$\cos(\theta)\theta'=1,$$
 +
or equivalently
 +
$$\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{\cos(\theta)}.$$
 
The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$:
 
The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$:
 
[[File:Cos(arcsin(z)).png|200px|center]]
 
[[File:Cos(arcsin(z)).png|200px|center]]
Line 16: Line 18:
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 +
[[Category:Proven]]

Latest revision as of 23:33, 8 December 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}},$$ where $\arcsin$ denotes the inverse sine function.

Proof

If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use implicit differentiation with respect to $z$ and the derivative of sine to get $$\cos(\theta)\theta'=1,$$ or equivalently $$\dfrac{\mathrm{d}\theta}{\mathrm{d}z} = \dfrac{1}{\cos(\theta)}.$$ The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$:

Cos(arcsin(z)).png

Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}},$$ as was to be shown. █

References