Can anyone tell me what is the derivative of sin^2(3x + 8)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the derivative of [sin(3x + 8)]^2.

Use the chain rule.

[(sin(3x + 8))^2]'

=> 2*sin(3x + 8)*[sin (3x + 8)]'*[3x + 8]'

=> 2*sin(3x + 8)*cos (3x + 8)*3

=> 6*sin(3x + 8)*cos (3x + 8)

The derivative of [sin(3x + 8)]^2 is 6*sin(3x + 8)*cos (3x + 8)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

You'll have to use chain rule to determine the 1st derivative of the function:

f(g(x)) = f'(x)*g'(x)*x'

y' = 2sin (3x+8)*[cos(3x+8)]*[(3x+8)']

y' = 6sin(3x+8)*[cos(3x+8)]

The derivative of the function is y' = 6sin(3x+8)*[cos(3x+8)]

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