# Find the First Two Derivatives of 2 sinx cosx.

The derivative of a function is represented by or f ‘(x). It means that the function is the derivative of y with respect to the variable x. It is calculated using the method of differentiation.

## Answer: The first two derivatives of 2 sinx cosx are 2cos(2x) and -4sin(2x) respectively

We will use different trigonometric formulas to calculate the derivatives.

**Explanation:**

To find first derivative:

Let y = 2sin(x)cos(x)..............(1)

Now, Using the product rule:on (1)

uv’ + vu’

Here u is 2sin(x) and v is cos(x)

Now, y’ = 2sin(x)(-sin(x)) + cos(x)2cos(x)

On simplifying we get,

y’ = 2cos^{2}(x)-2sin^{2}(x)

y’ = 2(cos^{2}(x) - sin^{2}(x))

Also, cos(2x) = cos^{2}(x)-sin^{2}(x)

So, after substituting it in the above equation,

We get, y’ = 2cos(2x)

Hence, First derivative of 2sin(x)cos(x) is 2cos(2x)

Now will find second derivative of y = 2sin(x)cos(x) using chain rule that is fond the derivateive of y’ = 2cos(2x).

y” = 2(-sin(2x) 2)

y” = -4sin(2x)

### Thus, the first two derivatives of 2 sinx cosx are 2cos(2x) and -4sin(2x) respectively

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