Question

doc d 1. Using Implicit differentiation derive the formula -(arcsin x) V1 - 22 2. Find the equation of tangent to the curve y = arccos(.x3 – 1) at the point (1,0).

Answer 1

solution Y = grosinca we know y = arcsinlay is inverge pynation OF Y = sina le Y = gresima =). x= siny using x= siny we need to differentiate with mespact to x, So this will the be differentiate implicity need Remembering that a (Argy dze & (Figs. dy x= sing In Cala of any coing) day dy 1) Cosy 1 are =) dy 2 Cuៅ។ an Y = are simlas 1 Coslare singl

sinto two-1 coso = + JI-sinry me Here we take only positive root (reason is given in volej ㅛ Cas (are sing) Ji-sim?( are sinton] so dy da T Cos (are sin(a) Vise pote Reason why we take only the cosa= J I-sizzy (here) 4 = sin(x) how an inverse enly if we nestrict to domain to Isxstila " 9 10 am angle and it 12 conly google cosy in the namye - Laath - In

This range is in the I and I quadrante where the cosine safio are positive T/2 I -Т, lo are cos (x²+) Given cuore YP and point (20) we know that the equation of tamperit line at a point (x1,9) By 4-9, = mla-W en q curve 13 given where mß slope at لولا ,x) m = da

So = are cos are cos (x² V h37 cosy. - siny a la 3x2 11 dy dre 3 x2 ㄲ siny sinly tworzy 2 siny= sin cosy cli- so dy da 3x ? 17 1 - 42 Now at (20) slope dy -3(272 m = da 112,0) = all JI- (2-13² - 3 So apaolin of tangent line Y-0= -3(x-2) 7 + 3x = 3