Question 1

A) The derivative of f(x) = 6x ^ 2 is given by f^ prime (x)=lim h——>0 ________=____.

B) The derivative of f(x) = 2x ^ 2 - 7x + 8 is given by f () lim h—->______=____.

Question 2

Here we go ~

A.) Derivative of 6x² :

$\qquad \sf  \dashrightarrow \:f {}^( \prime)(x) = \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: (f(x + h) - f(x))/(h)$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{6(x + h) {}^(2) - 6(x) {}^(2) }{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{6(x {}^(2) + 2xh + h {}^(2) ) {}^{} - 6(x) {}^(2) }{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ \cancel{6x {}^(2)} + 12xh +6 h {}^(2) {}^{} - \cancel{6x{}^(2)} }{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ 12xh +6 h {}^(2) {}^{} }{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ \cancel{ h}( 12x +6h ) {}^{} {}^{} }{ \cancel{h}}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: 12x + 6h$

$\qquad \sf  \dashrightarrow \: 12x + 0$

$\qquad \sf  \dashrightarrow \: f {}^( \prime) (x) = 12x$

B.) The derivative of f(x) = 2x² -7x + 8 :

$\qquad \sf  \dashrightarrow \:f {}^( \prime)(x) = \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: (f(x + h) - f(x))/(h)$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ 2(x + h) {}^(2) - 7(x + h) + 8 - (2 {x}^(2) - 7x + 8)}{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ 2(x {}^(2) + 2xh + {h}^(2) ) {}^{} - \cancel{7x} + 7h+ \cancel8 - 2 {x}^(2) + \cancel{7x } - \cancel 8)}{h}$

$\qquad \sf  \dashrightarrow \: \sf\displaystyle { \lim_(h\to0)} \sf\: \: \frac{ \cancel{ 2x {}^(2)} + 4xh + 2{h}^(2) {}^{} - 7h - \cancel{2 {x}^(2)} }{h}$