Inverse Trig Functions

6Chain Rule VersionsThe chain rule versions of both derivative formulas are:ddx (arcsin u) = 1 du√1 − u 2 dxd1 du(arctan u) =dx 1 + u 2 dxEXAMPLE 24.7. Let’s use these formulas to find the derivatives of the following:ddx (arctan 1e3x ) =1 + (e 3x ) 2 · 3e3x = 3e3x1 + e 6x . (u = e3x )ddx (arcsin 3x2 ) =1√1 − (3x 2 ) 2 · 6x =6x√1 − 9x 4 . (u = 3x2 )ddx (earctan 3x ) = e arctan 3x 1 3earctan3x· 3 =1 + 9x2 1 + 9x 2 .ddx (sin 2x arctan 5x2 ) = 2 cos 2x arctan 5x 2 + sin 2x ·= 2 cos 2x arctan 5x 2 +10x sin 2x1 + 25x 4 .11 + 25x 4 · 10xddx (ln | arcsin 3x|) = 1arcsin 3x · 1√ · 3 = 31 − 9x 2 (arcsin 3x) √ 1 − 9x . 2D x (| arcsin(ln 3x)) =1√ · 11 − [ln(3x)] 2 3x · 3 = 1x √ 1 − [ln(3x)] . 2(u = ln(3x))YOU TRY IT 24.3. Find the derivatives of these functions:(a)dddx arctan(6x2 )] (b)dx [arcsin(√ x)](e)ddx [arctan(ln |6x|)] (f ) ddx [arcsin(6esin x )](i)ddx (ln | arctan +1 ex4 |)The answers are on the next page.(c)ddx [arctan(e2x ](g)ddx (e2 arcsin x2 )(d)d[arcsin(arcsin x)]dx(h)ddx [(arcsin 2x)(tan 5x2 )]