Question

What is the derivative of the function Y =X² Sin (x/2) Cos (x/2)=?

Answer 1

The derivative of the function
`\( Y = X^2 \sin\left((x)/(2)\) \cos\left((x)/(2)\right) \)` is
`\( Y' = 2X\sin\left((x)/(2)\)\cos\left((x)/(2)\right) + X^2\left((\cos(x))/(2) - \sin(x)\right) \).`

Step-by-step explanation:

To find the derivative of the given function
`\( Y = X^2 \sin\left((x)/(2)\) \cos\left((x)/(2)\right) \),` we apply the product rule. The product rule states that if
`\( Y = uv \),` where ( u ) and ( v ) are functions of ( X ), then
`\( Y' = u'v + uv' \).` Here,
`\( u = X^2 \)` and
`\( v = \sin\left((x)/(2)\) \cos\left((x)/(2)\right) \).` Calculating the derivatives of ( u ) and ( v ) and applying the product rule yields the expression for the derivative:

`\[ Y' = 2X\sin\left((x)/(2)\)\cos\left((x)/(2)\right) + X^2\left((\cos(x))/(2) - \sin(x)\right) \]`

Breaking down the components of the expression, the first term
`\( 2X\sin\left((x)/(2)\)\cos\left((x)/(2)\right) \)` represents the derivative of the quadratic term
`\( X^2 \)`multiplied by the original trigonometric function. The second term
`\( X^2\left((\cos(x))/(2) - \sin(x)\right) \)` results from the product of the original quadratic term
`\( X^2 \)` and the derivative of the trigonometric function
`\( \sin\left((x)/(2)\)\cos\left((x)/(2)\right) \).` Therefore, the derivative ( Y' ) captures the rate of change of the given function with respect to ( X ).