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

sarkisss454 sarkisss454

16-09-2022
Mathematics

Answered

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—->=.

Answer :

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DᴀʀᴋPᴀʀᴀᴅᴏx DᴀʀᴋPᴀʀᴀᴅᴏx · Answer 1 · 2022-09-17T00:09:29-04:00

Here we go ~

A.) Derivative of 6x² :

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

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

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

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel{6x {}^{2}} + 12xh +6 h {}^{2} {}^{} - \cancel{6x{}^{2}} }{h} [/tex]

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ 12xh +6 h {}^{2} {}^{} }{h} [/tex]

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

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: 12x + 6h[/tex]

[tex]\qquad \sf \dashrightarrow \: 12x + 0[/tex]

[tex]\qquad \sf \dashrightarrow \: f {}^{ \prime} (x) = 12x[/tex]

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

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

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

[tex]\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} [/tex]

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel{ 2x {}^{2}} + 4xh + 2{h}^{2} {}^{} - 7h - \cancel{2 {x}^{2}} }{h} [/tex]

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: \frac{ \cancel {h}( 4x + 2{h}^{} {}^{} - 7) }{ \cancel{h} }[/tex]

[tex]\qquad \sf \dashrightarrow \: \sf\displaystyle { \lim_{h\to0}} \sf\: \: 4x + 2{h}^{} {}^{} - 7[/tex]

[tex]\qquad \sf \dashrightarrow \: 4x - 0 - 7[/tex]

[tex]\qquad \sf \dashrightarrow \: f {}^{ \prime} (x) = 4x - 7[/tex]

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—->______=____.

Answer :

Other Questions

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—->=.