Differentiate xcos2x.derivative of xcos2x is =?

We have to differentiate f(x) = x*cos 2x f'(x) = x'*cos 2x + x*(cos 2x)' f'(x) = cos 2x + x*(-sin 2x)*2 f'(x) = cos 2x - 2x*(sin 2x) The required derivative of f(x) = x*cos 2x is f'(x) = cos 2x - 2x*(sin 2x)

Let y = x*cos2x We'll apply the chain rule and also the product rule. dy/dx = [d/dx(x)]* cos2x + x* [d/dx (cos 2x)] We'll apply the chain rule for the term: [d/dx (cos 2x)] = -sin 2x* (2x)' [d/dx (cos 2x)] = -2sin 2x The differentiated given function is: dy/dx = cos 2x - 2x(sin 2x)

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Differentiate xcos2x.derivative of xcos2x is =?

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2 Answers

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on May 11, 2011 at 6:34 AM

We have to differentiate f(x) = x*cos 2x

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

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

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

The required derivative of f(x) = x*cos 2x is f'(x) = cos 2x - 2x*(sin 2x)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on May 10, 2011 at 11:00 AM

Let y = x*cos2x

We'll apply the chain rule and also the product rule.

dy/dx = [d/dx(x)]* cos2x + x* [d/dx (cos 2x)]

We'll apply the chain rule for the term:

[d/dx (cos 2x)] = -sin 2x* (2x)'

[d/dx (cos 2x)] = -2sin 2x

The differentiated given function is: dy/dx = cos 2x - 2x(sin 2x)