[tex]\begin{gathered} f(x)=3x^3-7x-5 \\ f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h} \\ f^{\prime}(x)=\lim _{h\to0}\frac{3(x+h)^3-7(x+h)-5-(3x^3-7x-5)}{h} \\ (x+h)^3=x^3+3x^2h+3xh^2+h^3 \\ f^{\prime}(x)=\lim _{h\to0}\frac{3(x^3+3x^2h+3xh^2+h^3)^{}-7(x+h)-5-(3x^3-7x-5)}{h} \\ \\ f^{\prime}(x)=\lim _{h\to0}\frac{3x^3+9x^2h+9xh^2+3h^3-7x-7h-5-3x^3+7x+5}{h} \\ f^{\prime}(x)=\lim _{h\to0}\frac{(3x^3-3x^3)+9x^2h+9xh^2+3h^3+(-7x+7x)-7h+(-5+5)}{h} \\ f^{\prime}(x)=\lim _{h\to0}\frac{9x^2h+9xh^2+3h^3-7h}{h} \\ f^{\prime}(x)=\lim _{h\to0}\frac{h(9x^2+9xh^{}+3h^2-7)}{h} \\ f^{\prime}(x)=\lim _{h\to0}9x^2+9xh^{}+3h^2-7=9x^2+9x(0)+3(0)^2-7=9x^2-7 \\ f^{\prime}(x)=9x^2-7 \\ The\text{ first derivative is }9x^2-7 \\ For\text{ f''(x)} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{f^{\prime}(x+h)-f^{\prime}(x)}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{9(x+h)^2-7-(9x^2-7)}{h} \\ (x+h)^2=x^2+2xh+h^2 \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{9(x^2+2xh+h^2)^{}-7-(9x^2-7)}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{9x^2+18xh+9h^2^{}-7-9x^2+7}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{(9x^2-9x^2)+18xh+9h^2+(-7+7)}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{18xh+9h^2}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}\frac{h(18x+9h)}{h} \\ f^{\prime\prime}(x)=\lim _{h\to0}18x+9h=18x+9(0)=18x \\ f^{\prime\prime}(x)=18x \\ \text{The second derivative is 18x} \end{gathered}[/tex]
