What is the derivative of : y=sin(ln(sqrt(6x^2*lnx^2)))

Asked on by axel12

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We can differentiate y=sin(ln(sqrt(6x^2*lnx^2))) by using the chain rule.


y' = cos(ln(sqrt(6x^2*lnx^2)))*(1/(sqrt(6x^2*lnx^2)))*(1/2)*(1/sqrt(6x^2*lnx^2))*(6x^2*2x/x^2 + 12x*lnx^2)

y'= cos(ln(sqrt(6x^2*lnx^2)))*(1/(sqrt(6x^2*lnx^2)))*(1/2)*(1/sqrt(6x^2*lnx^2))*(12x + 12x*lnx^2)

y'= cos(ln(sqrt(6x^2*lnx^2)))*(1/(6x^2*lnx^2))*(6x + 6x*lnx^2)

y'= cos(ln(sqrt(6x^2*lnx^2)))*(1/(x*lnx^2))*(1 + lnx^2)

The required derivative is: y'= cos(ln(sqrt(6x^2*lnx^2)))*(1/(x*lnx^2))*(1 + lnx^2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since the given function is the result of composition of more than 2 functions, we'll apply chain rule:

y' = [sin(ln(sqrt(6x^2*lnx^2)))]'*(ln(sqrt(6x^2*lnx^2)))'*(sqrt(6x^2*lnx^2))'*(6x^2*lnx^2)'

The last factor has to be differentiated applying the product rule and chain rule.

y' = [cos(ln(sqrt(6x^2*lnx^2)))]*[1/(sqrt(6x^2*lnx^2))]*[1/2*sqrt(6x^2*lnx^2)]*[(6x^2)'*lnx^2 + (6x^2)*(lnx^2)']

y' = [cos(ln(sqrt(6x^2*lnx^2))]*(12x*ln x^2 + 6x^2*2x/x^2)/2*(6x^2*lnx^2)

y' = 12x*(1+lnx^2)*[cos(ln(sqrt(6x^2*lnx^2))]/12x^2*lnx^2

y' = (1+lnx^2)*[cos(ln(sqrt(6x^2*lnx^2))]/x*lnx^2

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