# Find the derivative of y =11^(x)sqrt(1-x)(sin(3x))^(4)/((x^2 + 1)^(3)*x^(1/5)). You need to differentiate the function with respect to x using chain rule, hence you should start to differentiate from outside to inside such that:

`y'= 11^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5))) ln 11* ^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))'`

You need to differentiate using quotient rule such that:

`(((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))' = ((sqrt(1-x)sin(3x))^4)'*(x^2 + 1)^3*x^(1/5)...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

You need to differentiate the function with respect to x using chain rule, hence you should start to differentiate from outside to inside such that:

`y'= 11^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5))) ln 11* ^((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))'`

You need to differentiate using quotient rule such that:

`(((sqrt(1-x)(sin(3x))^4)/((x^2 + 1)^(3)*x^(1/5)))' = ((sqrt(1-x)sin(3x))^4)'*(x^2 + 1)^3*x^(1/5) - sqrt(1-x)(sin(3x))^4*((x^2 + 1)^3*x^(1/5))')/((x^2 + 1)^(6)*x^(2/5))`

You need to differentiate each product from numerator using product rule such that:

`(((sqrt(1-x)(sin(3x))^4))'*((x^2 + 1)^(3)*x^(1/5))) = (-(sin 3x)^4/(2sqrt(1-x)) + 12sqrt(1-x)*(sin 3x)^3*(cos 3x))*((x^2 + 1)^(3)*x^(1/5)))`

`(((sqrt(1-x)(sin(3x))^4))*((x^2 + 1)^(3)*x^(1/5))') = (((sqrt(1-x)(sin(3x))^4))*((6x^(6/5)*(x^2+1)^2 + (1/5)(x^2 + 1)^(3)*x^(-4/5))`

Hence, differentiating with respect to x yields:

`y' = (((-(sin 3x)^4/(2sqrt(1-x)) + 12sqrt(1-x)*(sin 3x)^3*(cos 3x))*(x^2 + 1)^(3)*x^(1/5)) - (sqrt(1-x)sin(3x)^4)*(6x^(6/5)*(x^2+1)^2 + (1/5)(x^2 + 1)^3*x^(-4/5)))/((x^2 + 1)^(6)*x^(2/5)))`

Approved by eNotes Editorial Team