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Mnenie [13.5K]
3 years ago
5

Derivative of y=cos²x-sin²x

Mathematics
1 answer:
Leona [35]3 years ago
8 0
One thing to bear in mind with trigonometric expressions is that, the exponent location is next to the function name, instead of the whole expression, though it applies to the whole thing, namely cos²(x), is really [ cos(x) ]², and that matters for using the chain rule.

\bf y=cos^2(x)-sin^2(x)\implies y=[cos(x)]^2-[sin(x)]^2
\\\\\\
\cfrac{dy}{dx}=\stackrel{chain~rule}{2cos(x)\cdot -sin(x)}~~-~~\stackrel{chain~rule}{2sin(x)\cdot cos(x)}
\\\\\\
\cfrac{dy}{dx}=-2cos(x)sin(x)-2cos(x)sin(x)
\\\\\\
\cfrac{dy}{dx}=-4cos(x)sin(x)
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17(w-4) +3w -27 <br><br><br> Helpppo
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Answer:

w = 19/4

Step-by-step explanation:

Solve for w:

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17 (w - 4) = 17 w - 68:

17 w - 68 + 3 w - 27 = 0

Grouping like terms, 17 w + 3 w - 68 - 27 = (17 w + 3 w) + (-27 - 68):

(17 w + 3 w) + (-27 - 68) = 0

17 w + 3 w = 20 w:

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Add 95 to both sides:

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95 - 95 = 0:

20 w = 95

Divide both sides of 20 w = 95 by 20:

(20 w)/20 = 95/20

20/20 = 1:

w = 95/20

The gcd of 95 and 20 is 5, so 95/20 = (5×19)/(5×4) = 5/5×19/4 = 19/4:

Answer: w = 19/4

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