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You can actually use either the product rule or the chain rule for this one. Observe:
• Method I:y = cos² xy = cos x · cos xDifferentiate it by applying the product rule:

The derivative of
cos x is
– sin x. So you have


—————
• Method II:You can also treat
y as a composite function:

and then, differentiate
y by applying the chain rule:

For that first derivative with respect to
u, just use the power rule, then you have

and then you get the same answer:

I hope this helps. =)
Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>
Answer:
<h3>m∠1 = 41°</h3><h3>m∠2 = 41°</h3>
Step-by-step explanation:
<h3>m∠1 = m∠2</h3>

x = 5°,
m∠1 = (6x + 11)°

Answer:
5c-9h
Step-by-step explanation:
1) 7c-4h-2c-5h
2) 5c-4h-5h
3) 5c-9h
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