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y = 2sin(x)cos(x)
Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:
y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)
Simplify:
y' = 2cos2(x)-2sin2(x)
y' = 2(cos2(x)-sin2(x))
Use trig identity cos(2x) = cos2(x)-sin2(x):
y' = 2cos(2x)
Take second derivative using chain rule:
y'' = 2(-sin(2x)cos(2x))
Simplify:
y'' = -2sin(2x)(2)
Simplify:
y'' = -4sin(2x)y'' = -4sin(2x)
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∙ 11y ago
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What is the anti-derivative of -2cosx?
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How would you find x when 0 equals 2sinxcosx-cosx?
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What is the derivative of -cosx?
-sinx
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Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
How do you take the derivative of a trig function?
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
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d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
What is the differentiation of sin x over 1 plus cosx?
The derivative is 1/(1 + cosx)
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derivative (7cosx) = -ln(7) 7cosx sinx dx
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Find the derivative of sinxcosx?
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