## University Calculus: Early Transcendentals (3rd Edition)

$$P'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+2a_2x+a_1$$
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+ ... + a_2x^2+a_1x^1+a_0$$ where $a_n\ne0$ We find $P'(x)$: $$P'(x)=(a_nx^n)'+(a_{n-1}x^{n-1})'+ ... + (a_2x^2)'+(a_1x^1)'+(a_0)'$$ We will apply the Derivative Product Rule here: $$P'(x)=\Big((a_n)'x^n+a_n(x^n)'\Big)+\Big((a_{n-1})'x^{n-1}+a_{n-1}(x^{n-1})'\Big)+...+\Big((a_2)'x^2+a_2(x^2)'\Big)+\Big((a_1)'x+a_1(x)'\Big)+0$$ $$P'(x)=\Big(0\times x^n+a_n(nx^{n-1})\Big)+\Big(0\times x^{n-1}+a_{n-1}(n-1)x^{n-2}\Big)+...+\Big(0\times x^2+a_2(2x)\Big)+\Big(0\times x+a_1\times1\Big)$$ $$P'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+2a_2x+a_1$$