How to find d/dx (sin⁻¹ x) without using θ (or any other variable) ∵ sin(sin⁻¹ x) = x ∴ d/dx sin(sin⁻¹ x) = 1 Differentiating: ⇒ cos(sin⁻¹ x)⋅d/dx (sin⁻¹ x) = 1 ⇒ d/dx (sin⁻¹ x) = 1/cos(sin⁻¹ x) ⇒ d/dx (sin⁻¹ x) = 1/√(1–sin²(sin⁻¹ x)) ⇒ d/dx (sin⁻¹ x) = 1/√(1–x²)
2026-09-05 16:54
@sarin_5063 27일 전
The reason why cos theta = sqrt(1-x^2) and not +-sqrt(1-x^2):
When you define theta = arcsin x, the range of possible values of theta becomes [-pi/2, pi/2]. For any value of theta in this range, cos theta >= 0. Therefore, the possibility that cos theta = -sqrt(1-x^2) is ruled out.
(This is also the same reason as to why you do not use absolute values when doing trig subs in integration)
2026-09-05 16:54
@crankykransky2… 27일 전
Now do it by first principles
2026-09-05 16:13
@OyatilloToshma… 27일 전
Day 121?
2026-09-05 16:07
@surenazand 27일 전
Thank you I learned so much ❤
2026-09-06 03:16
@Antonino-f1i 27일 전
Perfect ???
2026-09-06 02:43
@_zelatrix 27일 전
What's stopping me saying that 1/cos x = sec x at the end?
2026-09-05 21:51
@tionggeegoh986… 24일 전
this differentiation is only for |x|<½π hence cos theta >0 ie cos theta =√(1-x²)