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How do you differentiate # y = 3x cos (x/3) - sin (x/3)#?

Answer 1

Scarlett Allison

the answer
#y'=[8cos(x/3)]/3-xsin(x/3)#

show below

#y = 3x cos (x/3) - sin (x/3)#

#y'=3[x*-1/3sin(x/3)+cos(x/3)*1]-1/3*cos(x/3)#

#y'=-x*sin(x/3)+3cos(x/3)-1/3*cos(x/3)#

#y'=[8cos(x/3)]/3-xsin(x/3)#

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Answer 2

Isaiah Allen

To differentiate the given function ( y = 3x \cos \left(\frac{x}{3}\right) - \sin \left(\frac{x}{3}\right) ), apply the product rule and the chain rule:

[ y' = 3 \cos \left(\frac{x}{3}\right) - x \sin \left(\frac{x}{3}\right) \left(\frac{1}{3}\right) - \cos \left(\frac{x}{3}\right) ]

Simplify to get the final derivative:

[ y' = 3 \cos \left(\frac{x}{3}\right) - \frac{x}{3} \sin \left(\frac{x}{3}\right) - \cos \left(\frac{x}{3}\right) ]

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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.