Differentiate f(x) = 2sin(x^2)*cos(x^2).Differentiate f(x) = 2sin(x^2)*cos(x^2).

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function f(x) = 2sin(x^2)*cos(x^2) has to be differentiated. We use the product rule and the chain rule.

f'(x) = 2sin(x^2)'*cos(x^2) + 2sin(x^2)*cos(x^2)'

=> f'(x) = 2 cos (x^2)*2x*cos(x^2) - 2sin(x^2)*sin (x^2)*2x

=> f'(x) = 4x (cos (x^2))^2 - 4x (sin (x^2))^2

=> f'(x) = 4x (cos 2(x^2))

The derivative of f(x) = 2sin(x^2)*cos(x^2) is 4x*(cos 2(x^2))

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

f'(x) = [2sin(x^2)*cos(x^2)]'

We'll use the product rule and the chain rule:

(u*v)' = u'*v + u*v'

We'll put u(x) = 2sin(x^2) => u'(x) = 2(cos x^2)*(x^2)'

u'(x) = 4x*cos x^2

v(x) = cos(x^2) => v'(x) = [-sin(x^2)]*(x^2)'

v'(x) = -2x*sin(x^2)

We'll substitute f,g,f',g' in the expression of product:

f'(x) = 4x*cos x^2*cos x^2 - 2sin(x^2)*2x*sin(x^2)

f'(x) = 4x(cos x^2)^2 - 4x(sin x^2)^2

We'll factorize by 4x:

f'(x) = 4x[(cos x^2)^2 - (sin x^2)^2]

f'(x) = 4xcos (2x^2)

We’ve answered 318,960 questions. We can answer yours, too.

Ask a question