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How do you differentiate #f(x)=(x+sinx)(x-cosx)# using the product rule?

Answer 1

William Abdalla

#f=x+sinx, g=x-cosx; ---- >f'=1+cosx, g'=1+sinx#
#f'(x)=fg'+gf'=(x+sinx)(1+sinx)+(x-cosx)(1+cosx)#

Divide the products into f and g, then determine each one's derivative independently and add it to the product rule.

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Answer 2

Axel Alvarez

To differentiate the function ( f(x) = (x + \sin x)(x - \cos x) ) using the product rule, we let ( u = x + \sin x ) and ( v = x - \cos x ). Then, applying the product rule ( (uv)' = u'v + uv' ), we find:

( u' = 1 + \cos x ) and ( v' = 1 + \sin x ).

Substituting these values into the product rule formula, we get:

( f'(x) = (1 + \cos x)(x - \cos x) + (x + \sin x)(1 - \sin x) )

Expanding and simplifying:

( f'(x) = x - x \cos x + \cos x - \cos^2 x + x - x \sin x + \sin x - \sin^2 x )

( f'(x) = x - x \cos x + x - x \sin x )

( f'(x) = x - x(\cos x + \sin x) )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.