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How to find dy/dx if y=ln(sin(3x^2-1))*ln(cos(3x^2-1))?

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sauerrahm | Student | eNotes Newbie

Posted May 7, 2011 at 10:49 PM via web

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How to find dy/dx if y=ln(sin(3x^2-1))*ln(cos(3x^2-1))?

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giorgiana1976 | College Teacher | Valedictorian

Posted May 7, 2011 at 10:58 PM (Answer #1)

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To determine dy/dx, we'll have to use two rules of differentiation: chain rule and product rule.

The product rule states:

(f*g)' = f'*g + f*g'

The chain rule states;

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

We'll write the product rule first:

dy/dx = [ln(sin(3x^2-1))]'*ln(cos(3x^2-1)) + ln(sin(3x^2-1))*[ln(cos(3x^2-1))]'

Now, we'll apply the chain rule:

dy/dx = 6x*cos(3x^2-1)*ln(cos(3x^2-1))/(sin(3x^2-1) -  6x*sin(3x^2-1)*ln(sin(3x^2-1))/cos(3x^2-1)

dy/dx = 6x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) - [sin(3x^2-1)]^2*ln(sin(3x^2-1))}/sin(3x^2-1)*cos(3x^2-1)

dy/dx = 12x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) - [sin(3x^2-1)]^2*ln(sin(3x^2-1))}/2*sin(3x^2-1)*cos(3x^2-1)

We've created at denominator, the formula of sine of double angle:

dy/dx = 12x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) - [sin(3x^2-1)]^2*ln(sin(3x^2-1))}/ sin (6x^2 - 2)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 7, 2011 at 11:21 PM (Answer #2)

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We have to find the derivative of y=ln(sin(3x^2-1))*ln(cos(3x^2-1))

y = ln(sin (3x^2-1))*ln(cos (3x^2-1))

y' = [ln(sin (3x^2-1))]'*ln(cos (3x^2-1)) + ln(sin (3x^2-1))*[ln(cos (3x^2-1))]'

=> (1/sin(3x^2 - 1))*(cos (3x^2 - 1)*6x*ln(cos (3x^2-1)) - ln(sin (3x^2-1))*(1/cos(3x^2 - 1))*sin(3x^2 - 1)*6x

=> cot (3x^2 - 1)*6x*ln(cos (3x^2-1)) - ln(sin (3x^2-1))*(tan (3x^2 - 1)*6x

=> 6x*[cot (3x^2 - 1)*ln(cos (3x^2-1)) - ln(sin (3x^2-1))*(tan (3x^2 - 1)]

The required derivative dy/dx = 6x*[cot (3x^2 - 1)*ln(cos (3x^2-1)) - ln(sin (3x^2-1))*(tan (3x^2 - 1)]

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