Karl's Calculus Tutor - Solution to Exercise 7.2-2

Solution to Exercise 7.2-2

Step 1: You should be familiar enough with the product rule by now to apply it with little effort. If g(x) = sin(x) and h(x) = cos(x) and f(x) = g(x)h(x) = sin(x)cos(x), then you have

   f'(x)  =  g'(x)h(x) + g(x)h'(x)  =  cos(x)cos(x) - sin(x)sin(x)

You could simplify this still further to f'(x) = cos²(x) - sin²(x), but for the next part of the question, it will be useful to remember the unsimplified form. Now apply the chain rule to find the derivative of f(x) = (1/2)sin(2x). When you are done, click here.

Step 2: The derivative of the 2x that appears inside the sin function is simply 2, and the derivative of the sine is the cosine. So you should get

   f'(x)  =  (1/2)cos(2x) × 2  =  cos(2x)

Now, is there a trig identity that you can use that demonstrates that cos(2x) is the same as what you got when you took the derivative of sin(x)cos(x) using the product rule? See if you can figure out what it is and apply it. Then click here.

Step 3: Clearly cos(2x) = cos(x+x). Now you can use the identity, cos(a+b) = cos(a)cos(b) - sin(a)sin(b). In this case, it clearly demonstrates that

   cos(2x)  =  cos(x+x)  =  cos(x)cos(x) - sin(x)sin(x)

And isn't the right-hand expression exactly what you got when you took the derivative of sin(x)cos(x) using the product rule?

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