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Calculus, derivatives. Please help! Show work, if possible. Thanks! :)

Calculus, derivatives. Please help! Show work, if possible. Thanks! :)-example-1
User Kadeshpa
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1 Answer

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18 votes

Answer:

a. y=4

b. graph the given and y=4

Explanation:

The given is y= 8 sin(x) cos(x)

to get y' which is the slope of the tangent line we take the derivative

for this problem, the derivative involves the constant rule (which states that the derivative a constant times a value is just the constant times the derivative of said value) and the multiplication rule (assuming that the two parts in question are represented by the variables a and b then the derivative of a*b = a*b'+b*a')

thus we get 8*(sin(x)*-sin(x)+cos(x)*cos(x))

-8sin^2(x)+8cos^2(x)

You can also leave the 8 differentiated out depending on the form you need.

Now that we have the derivative we can plug pi/4 into this equation to get the slope of the tangent line at that point

=8sin^2(pi/4)+8cos^2(pi/4)

this would equal -8*1/2 +8*1/2 which equals 0

the slope of the tangent at this point would be zero

To get the point for point slope form we plug the pi/4 into the given equation to get a y value and get the point (pi/4, 4)

this means the point slope equation would be

y-4=0(x-pi/4)

y-4=0

y=4

To graph you would just enter the given equation and the equation for the tangent line into a graphing software or calculator and see if the line is tangent at the point (pi/4, 4)

User ThunderHorn
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