Calculus Differentiate Help?

Question

1. y = 4^x * sin x

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

Find dy/dx

Answer 1

y = 4^x × sin x

Diferenciate each side with respective x

dy /dx = d /dx (4^x × sin x)

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

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

f(x) = a^x then f¹(x) = a^xloga 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) = 4^x ,g(x) = sin x , f¹(x) = 4^xog4, g¹(x) = cos x and  y = f(x) g(x)

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

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

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

Answer 2

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(x^n-1) and d / dx (cscx) = -cscx cotx

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

dy /dx = -6csc¹(3x)[csc(3x) cot(3x)]

dy /dx = -6 csc²(3x) cot(3x)