Question

(2^sin^2(3x))'=
• A.
ln8 × 2^sin^2(3x) × sin6x
• B.
ln2× 2^sin^2(3x) × cos3x
• C.
e^sin^2(3x) × ln2 × cos^2(3x)
• D.
6cos3x × ln2 × 2 × sin^2(3x)

Answer 1

http://www.wolframalpha.com/input/?i=d%282%5Esin%5E2%283x%29%29%2Fdx

Answer 2

Let y = 2^sin^2 (3x).
Take the logarithm of both sides.
log_2 y = log_2 2^sin^2 (3x) = sin^2 3x
ln y / ln 2 = sin^2 3x
ln y = ln 2 sin^2 3x
Apply differentiation to both sides.
y' / y = 6 ln 2 sin 3x cos 3x = 3 ln 2 sin 6x
y' = 3y ln 2 sin 6x = 3 ln 2 sin 6x (2^sin^2 (3x))
Formulas and method used:
Logarithmic differentiation
log_a a^n = n
log_b a = log_c a / log_c b
2 sin a cos a = sin (2a)

Answer 3

Answer is A.
3 ln 2 sin 6x (2^sin^2 (3x)) = 2^sin^2 (3x) • ln 8 • sin 6x
Another formula used
a log_b c = log_b c^a

Answer 4

tHANKS for the reply