OpenStudy (anonymous):

find the derivative of sin[ln(cosx^3)]

4 years ago
OpenStudy (anonymous):

chain rule 3 times. lol

4 years ago
OpenStudy (directrix):

In sin[ln(cosx^3)] is (cosx^3) meant to be (cos(x)) ^3 OR is it meant to be ( cos(x^3) ) ? They are not the same.

4 years ago
OpenStudy (anonymous):

cos(x^3)

4 years ago
OpenStudy (anonymous):

here: we start by taking the derivative of the outside function sin(u) where u = ln(cosx^3). This is equal to cos(u)*du/dx. Then, du/dx requires the use of the chain rule again, therefore du/dx = d/dx ln(v) where v = cos(x^3). This is equal to (1/v)*dv/dx. We have that dv/dx requires use of the chain rule as well, so dv/dx = d/dx cos(k) where k = x^3. This is equal to -sin(k)*3x^2. So the final answer all together is: cos(ln(cosx^3))*(1/cosx^3)*(-sin(x^3))*3x^2

4 years ago
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