y=(1)/(4)x^4-(1)/(2)x^2+x+1 (dy)/(dx)=x^3-x+1

To find the derivative function \( y = \frac{1}{4}x^4 - \frac{1}{2}x^2 + x + 1 \), we will differentiate each term with respect to \( x \).<br /><br />Given:<br />\[ y = \frac{1}{4}x^4 - \frac{1}{2}x^2 + x + 1 \]<br /><br />We will differentiate each term separately:<br /><br />1. The derivative of \( \frac{1}{4}x^4 \):<br />\[ \frac{d}{dx} \left( \frac{1}{4}x^4 \right) = \frac{1}{4} \cdot 4x^3 = x^3 \]<br /><br />2. The derivative of \( -\frac{1}{2}x^2 \):<br />\[ \frac{d}{dx} \left( -\frac{1}{2}x^2 \right) = -\frac{1}{2} \cdot 2x = -x \]<br /><br />3. The derivative of \( x \):<br />\[ \frac{d}{dx} (x) = 1 \]<br /><br />4. The derivative of the constant term \( 1 \):<br />\[ \frac{d}{dx} (1) = 0 \]<br /><br />Now, we combine these results:<br />\[ \frac{dy}{dx} = x^3 - x + 1 \]<br /><br />So, the derivative of the function \( y = \frac{1}{4}x^4 - \frac{1}{2}x^2 + x + 1 \) is:<br />\[ \frac{dy}{dx} = x^3 - x + 1 \]