`y = x^2 sin(x) tan(x)` Differentiate.

Textbook Question

Chapter 3, 3.3 - Problem 16 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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y = x^2 sin(x) tan(x)

to find y'

so,

y' = (x^2 sin(x) tan(x))'

let a= sin(x) tan(x)

so,

y' = (x^2 * a)'

     = x^2 (a)'+ a *(2x)

where a' = ( sin(x) tan(x))'

             =  sin(x) d/(dx) (tan(x))+ tan(x) d/(dx)  (sin(x))

             = sin(x) * sec^2(X) + tan(x) * cos(x)

so, y' = x^2(sin(x) * sec^2(X) + tan(x) * cos(x)) +(sin(x) tan(x))(2x)

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