Trig inverse derivatives, further understanding

[YehiaMedhat]

Given that: [imath] \frac {d}{dx}f(x)= \frac {dy}{dx}[/imath] and [imath]\frac {d}{dx}f^-1(x)=\frac {dx}{dy}[/imath], therefore, if [imath]\frac {d}{dx}sin(x)=cos(x)[/imath], therefore, [imath]\frac {d}{dx}sin^-1(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}[/imath], but, in my lectures it was: [imath]\frac {d}{dx}sin^-1(x)=\frac {1}{\sqrt{1-x^2}}[/imath]
Is this similarity is my overthinking or there is a prove to how this [imath]\frac {d}{dx}sin^-1(x)=\frac {1}{\sqrt{1-x^2}}[/imath] came from [imath]\frac {d}{dx}sin^-1(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}[/imath]
Please, if you have any refrences, mention it.

 

[pka]

I seems that you may be confusing notation.
The inverse trig functions are often written as an exponent.
The inverse sin is often given as [imath]\sin^{-1}(x)\ne\dfrac{1}{\sin(x)}[/imath]
The correct notation is [imath]\bf\arcsin(x)[/imath] the derivative of which is [imath]\dfrac{1}{\sqrt{1-\sin^2(x)}[/imath]


[imath][/imath][imath][/imath]

[imath][/imath][imath][/imath]

 

[Dr.Peterson]

Given that: [imath] \frac {d}{dx}f(x)= \frac {dy}{dx}[/imath] and [imath]\frac {d}{dx}f^{-1}(x)=\frac {dx}{dy}[/imath], therefore, if [imath]\frac {d}{dx}sin(x)=cos(x)[/imath], therefore, [imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}[/imath], but, in my lectures it was: [imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{\sqrt{1-x^2}}[/imath]
Is this similarity is my overthinking or there is a prove to how this [imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{\sqrt{1-x^2}}[/imath] came from [imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos}=\frac {1}{\sqrt{1-sin^2(x)}}[/imath]
Please, if you have any refrences, mention it.

I think you do understand that [imath]\sin^{-1}(x)[/imath] means the inverse, not the reciprocal; the reciprocals you are taking are [intended to be] of the derivative. But you're doing that incorrectly, by taking it too fast.

Your main error is in saying "[imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos(x)}[/imath]", as if the latter were [imath]\frac {dx}{dy}[/imath].

Writing it out more carefully, if [imath]y=\sin^{-1}(x)[/imath], then [imath]x = \sin(y)[/imath], so that [imath]\frac{dx}{dy}=\cos(y)[/imath].

Now we can properly take the reciprocal: [math]\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\cos(y)}[/math]But this is in terms of y rather than x; so we need to express it in terms of x.

Since [imath]\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}[/imath], we can conclude that [math]\frac {d}{dx}\sin^{-1}(x)=\frac{dy}{dx}=\frac{1}{\cos(y)}=\frac{1}{\sqrt{1-x^2}}[/math]
(We can take the positive sign because the inverse sine is an angle whose cosine is positive.)

 

[YehiaMedhat]

I think you do understand that [imath]\sin^{-1}(x)[/imath] means the inverse, not the reciprocal; the reciprocals you are taking are [intended to be] of the derivative. But you're doing that incorrectly, by taking it too fast.

Your main error is in saying "[imath]\frac {d}{dx}sin^{-1}(x)=\frac {1}{cos(x)}[/imath]", as if the latter were [imath]\frac {dx}{dy}[/imath].

Writing it out more carefully, if [imath]y=\sin^{-1}(x)[/imath], then [imath]x = \sin(y)[/imath], so that [imath]\frac{dx}{dy}=\cos(y)[/imath].

Now we can properly take the reciprocal: [math]\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\cos(y)}[/math]But this is in terms of y rather than x; so we need to express it in terms of x.

Since [imath]\cos(y)=\sqrt{1-\sin^2(y)}=\sqrt{1-x^2}[/imath], we can conclude that [math]\frac {d}{dx}\sin^{-1}(x)=\frac{dy}{dx}=\frac{1}{\cos(y)}=\frac{1}{\sqrt{1-x^2}}[/math]
(We can take the positive sign because the inverse sine is an angle whose cosine is positive.)

Okay, I seem to have mistakingly written the formulai
Thanks for explaination