Find f''(x) in each of the following cases: f(x)=sin x sin 2x sin 3x

To find the second derivative, f''(x), of the function f(x) = sin(x) sin(2x) sin(3x), we will first need to compute the first derivative, f'(x), and then differentiate it again.

Step 1: First Derivative

Using the product rule, which states that the derivative of a product of functions is given by:

• (uv)' = u'v + uv'

Let:

• u = sin(x)
• v = sin(2x) sin(3x)

We first find the derivative of u:

• u' = cos(x)

Next, we need to differentiate v using the product rule again:

• v = sin(2x) * sin(3x)
• v' = cos(2x) * 2sin(3x) + sin(2x) * cos(3x) * 3

Now, applying the product rule to find f'(x):

• f'(x) = u'v + uv' = cos(x) * sin(2x) * sin(3x) + sin(x) * (cos(2x) * 2sin(3x) + sin(2x) * cos(3x) * 3)

Step 2: Second Derivative

Now, we differentiate f'(x) to find f''(x). This will involve applying the product rule and chain rule multiple times. The expression for f'(x) is quite complex, so we will differentiate each term carefully:

• Differentiate cos(x) * sin(2x) * sin(3x)
• Differentiate sin(x) * (cos(2x) * 2sin(3x) + sin(2x) * cos(3x) * 3)

After applying the product rule and simplifying, we will arrive at the final expression for f''(x).

Final Expression

The second derivative, f''(x), will be a combination of terms involving cos(x), sin(x), cos(2x), sin(2x), cos(3x), and sin(3x). The exact expression can be lengthy, but it will reflect the oscillatory nature of the original function.

In summary, to find f''(x), you first compute f'(x) using the product rule, and then differentiate f'(x) again, applying the product and chain rules as necessary.