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Chain Rule of Differentiation

Solve: y = ...

Question

Solve: $y = {sin}^{- 1} (cos x) + {cos}^{- 1} (sin x)$

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Solution

$y = {sin}^{- 1} (cos x) + {cos}^{- 1} (sin x)$
$y = \frac{π}{2} - {cos}^{- 1} (cos x) + \frac{π}{2} - {sin}^{- 1} (sin x)$
$y = \frac{π}{2} - x + \frac{π}{2} - x$
$y = π - 2 x$

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Similar questions

A

0 < x < \frac{π}{2}

{sin}^{- 1} (c o s x) + {cos}^{- 1} (s i n x) = π - 2 x

{cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x \forall x \in [0, 1]

y = | {cos}^{- 1} (sin x) | + ∣ ∣ ∣ \frac{π}{2} - ({cos}^{- 1} cos x) ∣ ∣ ∣, x

\frac{π}{2} < x < π

\frac{π^{2}}{k}

k

L e t f (x) = {s i n}^{- 1} s i n x + {c o s}^{- 1} c o s x,

g (x) = m x a n d h (x) = x

g (x)

f (x), x = π

y = 0

s i n^{- 1} (c o s (c o s^{- 1} (c o s x) + s i n^{- 1} (s i n x)))

x ϵ (\frac{π}{2}, π)

L e t f (x) = {s i n}^{- 1} (s i n x) + {c o s}^{- 1} (c o s x), g (x) = m x a n d h (x) = x

π

π

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