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Question

Find the first two derivatives of 2sinxcosx.


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Solution

Step 1: Compute the first derivative:

Suppose y=2sin(x)cos(x)

Now, we are using the product rule to find the derivatives.

The formula of the product rule:uv'+vu'

Here u=2sin(x) and v=cos(x).

Now, apply the product rule to find the first derivative,

y'=2sin(x)×ddx[cos(x)]+cos(x)×ddx[2sin(x)]y'=2sin(x)×[-sin(x)]+cos(x)×[2cos(x)]

Since the formula of ddx[cos(x)]=-sin(x) and ddx[sin(x)]=cos(x).

After simplifying the above expression we get,

y'=2[cos2x-sin2x]

y'=2cos(2x)cos2x=[cos2x-sin2x]

Step 2: Compute the second derivative:

Now, again apply the derivative on the first derivative and we get,

y"=2×ddx[cos(2x)]y"=-2×sin2x×ddx(2x)y"=-4×sin(2x)

Hence, the first two derivative of y=2sin(x)cos(x) is y"=-4sin(2x).


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