ind the derivative of the following f(x)=(1-5x^2)(6x^2-4x+1)

To find the derivative of the function $f(x)=(1-5x^{2})(6x^{2}-4x+1)$, we can use the product rule. The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by:<br /><br />$(uv)' = u'v + uv'$<br /><br />In this case, let $u(x) = (1-5x^{2})$ and $v(x) = (6x^{2}-4x+1)$. Then, we can find the derivatives of $u(x)$ and $v(x)$:<br /><br />$u'(x) = -10x$<br />$v'(x) = 12x - 4$<br /><br />Now, we can apply the product rule:<br /><br />$f'(x) = u'(x)v(x) + u(x)v'(x)$<br />$f'(x) = (-10x)(6x^{2}-4x+1) + (1-5x^{2})(12x-4)$<br /><br />Simplifying the expression, we get:<br /><br />$f'(x) = -60x^{3} + 40x^{2} - 10x + 12x - 48x + 4$<br />$f'(x) = -60x^{3} + 40x^{2} - 38x + 4$<br /><br />Therefore, the derivative of the function $f(x)=(1-5x^{2})(6x^{2}-4x+1)$ is $f'(x) = -60x^{3} + 40x^{2} - 38x + 4$.