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.
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’ = 2cos2(x)-2sin2(x)
y’ = 2(cos2(x) - sin2(x))
Also, cos(2x) = cos2(x)-sin2(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)