If y= sin 98(x) ⋅ cos 39(x), find (d y/d x)

Question

If y= sin 98(x) ⋅ cos 39(x), find (d y/d x) (A) (98 cos 99 x ⋅ sin 38 x)+(39 sin 40 x ⋅ cos 97 x) (B) (98 cos 98 x ⋅ sin 39 x)-(40 sin 39 x ⋅ cos 98 x) (C) (98 cos 99 x ⋅ sin 38 x)-(39 sin 40 x ⋅ cos 97 x) (D) (98 cos 98 x ⋅ sin 39 x)+(39 sin 40 x ⋅ cos 97 x)

Tardigrade Admin · Accepted Answer

y= sin 98 x ⋅ cos 39 x (d y/d x)= sin 98 x ⋅ 39 cos 38 x(- sin x)+ cos 39 x ⋅ 98 sin 97 x( cos x) =98 sin 97 x ⋅ cos 40 x-39 sin 99 x ⋅ cos 38 x =98 cos 99 x ⋅ sin 38 x-39 sin 40 x ⋅ cos 97 x

Q. If $y = sin^{98} (x) \cdot cos^{39} (x)$ , find $\frac{d y}{d x}$

$(98 cos^{99} x \cdot sin^{38} x) + (39 sin^{40} x \cdot cos^{97} x)$

$(98 cos^{98} x \cdot sin^{39} x) - (40 sin^{39} x \cdot cos^{98} x)$

$(98 cos^{99} x \cdot sin^{38} x) - (39 sin^{40} x \cdot cos^{97} x)$

$(98 cos^{98} x \cdot sin^{39} x) + (39 sin^{40} x \cdot cos^{97} x)$

$y = sin^{98} x \cdot cos^{39} x$
$\frac{d y}{d x} = sin^{98} x \cdot 39 cos^{38} x (- sin x) + cos^{39 x} \cdot 98 sin^{97} x (cos x)$
$= 98 sin^{97} x \cdot cos^{40} x - 39 sin^{99} x \cdot cos^{38} x$
$= 98 cos^{99} x \cdot sin^{38} x - 39 sin^{40} x \cdot cos^{97} x$

Q. If y=sin98(x)⋅cos39(x), find dxdy​

(98cos99x⋅sin38x)+(39sin40x⋅cos97x)

(98cos98x⋅sin39x)−(40sin39x⋅cos98x)

(98cos99x⋅sin38x)−(39sin40x⋅cos97x)

(98cos98x⋅sin39x)+(39sin40x⋅cos97x)