Question

Can someone help me solve this differentiation/tangent problem?

Answer 1

A)

$\bf g'(x)=\stackrel{product~rule}{2x\cdot f(x)+x^2\cdot f'(x)}\quad \begin{cases} x=5\\ f(5)=5\\ f'(5)=5 \end{cases} \\\\\\ g'(5)=2(5)\cdot f(5)+(5)^2\cdot f'(5)\implies g'(5)=50+500\\\\\\ g'(5)=550\\\\ -------------------------------\\\\ g(5)=(5)^2f(5)\implies g(5)=125 \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)} \begin{cases} x=5\\ y=125\\ \stackrel{m}{g'(5)}=550 \end{cases}\implies y-125=550(x-5) \\\\\\ y-125=550x-2750\implies y=550x+400$



b)

$\bf h'(x)=\stackrel{quotient~rule}{\cfrac{f'(x)(x-6)~~-~~f(x)\cdot 1}{(x-6)^2}}\quad \begin{cases} x=5\\ f(5)=5\\ f'(5)=5 \end{cases} \\\\\\ h'(5)=\cfrac{f'(5)(5-6)~~-~~f(5)\cdot 1}{(5-6)^2} \\\\\\ h'(5)=\cfrac{5(-1)-5}{(-1)^2}\implies h'(5)=-10\\\\ -------------------------------$

$\bf h(5)=\cfrac{f(5)}{5-6}\implies h(5)=\cfrac{5}{-1}\implies h(5)=-5 \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad \begin{cases} x=5\\ y=-5\\ \stackrel{m}{-10} \end{cases}\implies y-(-5)=-10(x-5) \\\\\\ y+5=-10x+50\implies y=-10x+45$