Question

2x³ differential coefficient​

Answer 1

Answer:

f'(x)=limh→0f(x+h)−f(x)h

f(x)=x3−2x2+x4+6

f(x+h)=(x+h)3−2(x+h)2+x+h4+6

f'(x)=limh→0(x+h)3−2(x+h)2+x+h4+6−(x3−2x2+x4+6)h

f'(x)=limh→0(x+h)3−2(x+h)2+x+h4+6−x3+2x2−x4−6h

f'(x)=limh→0(x+h)3−2(x+h)2+x+h4−x3+2x2−x4h

f'(x)=limh→0(x+h)3−2(x2+2xh+h2)+x+h4−x3+2x2−x4h

f'(x)=limh→0(x+h)3−2x2−4xh−2h2+x+h4−x3+2x2−x4h

f'(x)=limh→0(x+h)3−4xh−2h2+x+h4−x3−x4h

f'(x)=limh→0(x+h)(x2+2xh+h2)−4xh−2h2+x4+h4−x3−x4h

f'(x)=limh→0(x+h)(x2+2xh+h2)−4xh−2h2+h4−x3h

f'(x)=limh→0x3+2x2h+xh2+hx2+2xh2+h3−4xh−2h2+h4−x3h

f'(x)=limh→02x2h+xh2+hx2+2xh2+h3−4xh−2h2+h4h

f'(x)=limh→0h⋅(2x2+xh+x2+2xh+h2−4x−2h+14)h

f'(x)=limh→02x2+xh+x2+2xh+h2−4x−2h+14

f'(x)=2x2+x(0)+x2+2x(0)+(0)2−4x−2(0)+14

f'(x)=2x2+x2−4x+14

f'(x)=3x2−4x+14