Say we want to differentiate [itex]\arcsin x[/itex]. To do this we put [tex]y=\arcsin x.[/tex] Then [tex]x=\sin y \implies \frac{dx}{dy}= \cos y.[/tex] Then we use the relation [tex]\sin^2 y + \cos^2 y = 1 \implies \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}.[/tex] Therefore [tex]\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}.[/tex](adsbygoogle = window.adsbygoogle || []).push({});

My question is that when we use [tex]\sin^2 y + \cos^2 y = 1,[/tex] how do we know that [tex]\cos y = \sqrt{1 - \sin^2 y}[/tex] rather than [tex]\cos y = -\sqrt{1 - \sin^2 y}[/tex]?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Differentiation of inverse trig

Loading...

Similar Threads for Differentiation inverse trig |
---|

B When do we use which notation for Delta and Differentiation? |

B Product rule OR Partial differentiation |

A Differential operator, inverse thereof |

I Differentiation of sin function where's my mistake? |

I Differentials of order 2 or bigger that are equal to 0 |

**Physics Forums | Science Articles, Homework Help, Discussion**