# Calculus Differentiate Help?

1. y = 4^x * sin x

2. y = 1 / sin^2 (3x)

Find dy/dx
asked Mar 11, 2013 in CALCULUS

y = 4x × sin x

Diferenciate each side with respective x

dy /dx = d /dx (4x × sin x)

Put f(x) = 4x and g(x) = sin x

Diferenciate each side with respective x for f(x) and g(x)

f(x) = ax then f1(x) = axloga and g(x) = sin x then g1(x) = cos(x)

Therefore f1(x) = 4^x log4 and g1(x) = cos x

Recall : d /dx [f(x) g(x)] = f(x) g1(x) + f1(x) g(x)

Substitute f(x) = 4x ,g(x) = sin x , f1(x) = 4xog4, g1(x) = cos x and  y = f(x) g(x)

dy /dx = d /dx[4x ×sin x] = 4xcos x + 4xlog4 × sin x

dy /dx = d / dx[4x × sin x] = 4x[cos x + 2sin x log2]

Therefore dy /dx = 4x [cos x + 2sin x log2]

y = 1 / sin2 (3x)

Recall : csc2 (A) = 1 / sin2 (A)

y = csc2(3x)

Diferenciate each side with respective x

dy /dx = d / dx (csc2(3x)

Recall : d /dx (xn) = n(xn-1) and d / dx (cscx) = -cscx cotx

dy / dx = 2csc2-1(3x) 3 [-csc(3x) cot(3x)]

dy /dx = -6csc1(3x)[csc(3x) cot(3x)]

dy /dx = -6 csc2(3x) cot(3x)