Did I differentiate these two functions correctly?
y=2cos(2x)cos(14x)
y'=-14 sin(14x)2 cos(2x)-4 sin(2x)cos(14x)
y=2[cos 1/2(f1-f2 )x][cos 1/2(f1+f2 )x]
y^'=-[(f1+f2 )/2]sin[(f1+f2 )/2 x]2cos[(f1-f2 )/2 x] - [2(f1-f2 )/2]sin[(f1-f2 )/2 x]cos [(f1+f2)/2 x]
they appear correct
To determine if the differentiations were done correctly, we can compare the results to the usual rules of differentiation.
For the first function:
y = 2cos(2x)cos(14x)
To differentiate this function, we can apply the product rule. The product rule states that the derivative of a product of two functions is determined by taking the derivative of the first function multiplied by the second function, then adding to it the first function multiplied by the derivative of the second function.
Using the product rule, we have:
y' = (2cos(14x))(d/dx[cos(2x)]) + (d/dx[2cos(14x)])(cos(2x))
Now, let's differentiate each part separately:
d/dx[cos(2x)] = -2sin(2x)
d/dx[2cos(14x)] = -28sin(14x)
Plugging these back into the equation, we get:
y' = (2cos(14x))(-2sin(2x)) + (-28sin(14x))(cos(2x))
This simplifies to:
y' = -4sin(2x)cos(14x) - 28sin(14x)cos(2x)
From the calculation, we see that your differentiation for the first function is indeed correct.
For the second function:
y = 2[cos(1/2(f1-f2)x)][cos(1/2(f1+f2)x)]
Similarly, we can differentiate this function using the product rule. Let's break it down step by step:
First, let's rewrite the function as:
y = 2cos[(1/2)(f1-f2)x]cos[(1/2)(f1+f2)x]
Now, differentiating each part separately:
d/dx[cos[(1/2)(f1-f2)x]] = -[(f1-f2)/2]sin[(1/2)(f1-f2)x]
d/dx[cos[(1/2)(f1+f2)x]] = -[(f1+f2)/2]sin[(1/2)(f1+f2)x]
Plugging these back into the equation, we have:
y' = (2cos[(1/2)(f1-f2)x])(-[(f1+f2)/2]sin[(1/2)(f1+f2)x]) + ([-(f1-f2)/2]sin[(1/2)(f1-f2)x])(cos[(1/2)(f1+f2)x])
Simplifying further:
y' = -[(f1+f2)/2]sin[(f1+f2)/2 x]2cos[(f1-f2)/2 x] - [2(f1-f2)/2]sin[(f1-f2)/2 x]cos [(f1+f2)/2 x]
From the calculation, we can see that your differentiation for the second function is also correct.
In conclusion, both differentiations have been performed correctly based on the rules of differentiation.