# ` ` Q. If `y =` `(5x) / root(3) ( 1 - x^2)` `+` `sin^2(2x + 3)` . Find `dy/dx` .

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Given `y=(5x)/root(3)(1-x^2)+sin^2(2x+3)` we are asked to find `(dy)/(dx)` :

Rewrite as `y=5x(1-x^2)^(-1/3)+sin^2(2x+3)`

Use the product rule and the chain rule on the first term, and use the chain rule twice on the second term:

`y'=5x(-1/3)(1-x^2)^(-4/3)(-2x)+5(1-x^2)^(-1/3)`

`+2sin(2x+3)cos(2x+3)(2)`

`=10/3x^2(1-x^2)^(-4/3)+5(1-x^2)^(-1/3)`

`+4sin(2x+3)cos(2x+3)`

Factor out the common factor of `5(1-x^2)^(-4/3)` from the first two terms and use the double angle identity on the last term:

`=5(1-x^2)^(-4/3)[2/3x^2+1-x^2]+2sin(2(2x+3))`

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

or `(dy)/(dx)=-(5(x^2-3))/(3(1-x^2)^(4/3))+2sin(4x+6)`

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