## Calculus with Applications (10th Edition)

\begin{aligned} y=f(x) &=5 x^{2}-6 x+7 \\ y^{\prime} &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left[5(x+h)^{2}-6(x+h)+7\right]-\left[5 x^{2}-6 x+7\right]}{h} \\ &=\lim _{h \rightarrow 0} \frac{5\left(x^{2}+2 x h+h^{2}\right)-6 x-6 h+7-5 x^{2}+6 x+7}{h} \\ &=\lim _{h \rightarrow 0} \frac{10 x h+5 h^{2}-6 h}{h}=\lim _{h \rightarrow 0} \frac{h(10 x+5 h-6)}{h} \\ &=\lim _{h \rightarrow 0}(10 x+5 h-6)\\ &=10 x-6 \end{aligned} Thus the derivative of the given function is $$y^{\prime}=10 x-6$$
\begin{aligned} y=f(x) &=5 x^{2}-6 x+7 \\ y^{\prime} &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left[5(x+h)^{2}-6(x+h)+7\right]-\left[5 x^{2}-6 x+7\right]}{h} \\ &=\lim _{h \rightarrow 0} \frac{5\left(x^{2}+2 x h+h^{2}\right)-6 x-6 h+7-5 x^{2}+6 x+7}{h} \\ &=\lim _{h \rightarrow 0} \frac{10 x h+5 h^{2}-6 h}{h}=\lim _{h \rightarrow 0} \frac{h(10 x+5 h-6)}{h} \\ &=\lim _{h \rightarrow 0}(10 x+5 h-6)\\ &=10 x-6 \end{aligned} Thus the derivative of the given function is $$y^{\prime}=10 x-6$$