Question

How do I find the derivative of 2sinxcosx

Answer 1

$\large \large\begin{array}{l}{\textsf{Finding the derivative of}} \\\\ \mathsf{f(x)=2\,sin\,x\,cos\,x}\\\\\\ \bullet~~\textsf{Method \#1: Take derivatives using the product rule:}\\\\ \mathsf{f'(x)=(2\,sin\,x\,cos\,x)'}\\\\ \mathsf{f'(x)=(2\,sin\,x)'\cdot cos\,x+2\,sin\,x\cdot (cos\,x)'}\\\\ \mathsf{f'(x)=2\,cos\,x\cdot cos\,x+2\,sin\,x\cdot (-sin\,x)}\\\\ \mathsf{f'(x)=2\,cos^2\,x-2\,sin^2\,x}\\\\ \boxed{\begin{array}{c}\mathsf{f'(x)=2\cdot (cos^2\,x-sin^2\,x)} \end{array}}\qquad\checkmark \end{array}$


$\large \large\begin{array}{l} \bullet~~\textsf{Method \#2: Apply one of the double-angle trig formulas}\\\textsf{and use the chain rule:}\\\\ \mathsf{f(x)=2\,sin\,x\,cos\,x}\\\\ \mathsf{f(x)=sin\,2x}\\\\\\ \textsf{So,}\\\\ \mathsf{f'(x)=(sin\,2x)'}\\\\ \mathsf{f'(x)=cos\,2x\cdot (2x)'}\\\\ \mathsf{f'(x)=cos\,2x\cdot 2}\\\\ \boxed{\begin{array}{c}\mathsf{f'(x)=2\,cos\,2x} \end{array}}\qquad\checkmark \end{array}$


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Tags: derivative trig trigonometric function sine sin cosine cos product double angle chain rule differential calculus

Answer 2

( 2 sin x cos x )` = ( 2 sin x )` cos x + 2 sin x ( cos x )` =
= 2 cos x cos x + 2 sin x *( - sin x ) =
= 2 cos² x - 2 sin² x = 2 ( cos² x - sin² x ) = 2 cos ( 2 x )
Another way ( using trigonometric identities first ):
2 sin x cos x = sin ( 2 x )
( sin ( 2 x ) )` = 2 cos ( 2 x )