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How do you differentiate # y = sin 5x - 1/3 sin^3 5x#?

Answer 1

Scarlett Allison

#color(brown)(y' = 5 cos (5x) (1 - sin^2 (5x))#

#y = sin 5x - (1/3) sin^3 (5x)#

#y' = (dy)/(dx) = cos 5x * 5 - (1/cancel3) cancel3 sin^2 (5x) cos 5x * 5#

#y' = 5 cos (5x) - 5 sin^2 (5x) * cos (5x)#

#y' = 5 cos (5x) (1 - sin^2 (5x))#

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Answer 2

Emilia Aguilar

To differentiate ( y = \sin(5x) - \frac{1}{3} \sin^3(5x) ), we use the chain rule and the power rule for differentiation.

The derivative of ( \sin(5x) ) is ( 5 \cos(5x) ) by the chain rule, and the derivative of ( \sin^3(5x) ) requires the chain rule and the power rule: ( 3 \sin^2(5x) \cos(5x) ).

Therefore, the derivative of ( y ) is:

[ y' = 5 \cos(5x) - \frac{1}{3} \times 3 \sin^2(5x) \cos(5x) ]

Simplifying further:

[ y' = 5 \cos(5x) - \sin^2(5x) \cos(5x) ]

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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.