Solve this

Given: $f(x)=x^{3}+7 x^{2}+8 x-9$

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{(x+h)^{3}+7(x+h)^{2}+8(x+h)-9-x^{3}-7 x^{2}-8 x+9}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}+7 x^{2}+7 h^{2}+14 x h+8 x+8 h-9-x^{3}-7 x^{2}-8 x+9}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h^{3}+3 x^{2} h+3 x h^{2}+7 h^{2}+14 x h+8 h}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{h\left(h^{2}+3 x^{2}+3 x h+7 h+14 x+8\right)}{h}$

$\Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} h^{2}+3 x^{2}+3 x h+7 h+14 x+8$

$\Rightarrow f^{\prime}(x)=3 x^{2}+14 x+8$

Thus,

$f^{\prime}(4)=3 \times 4^{2}+14 \times 4+8$

$=48+56+8$

$=112$