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Use (a) the h→0definition of the derivative and then

(b) the z→cdefinition of the derivative to find f'(c) for each function f and value x=c in Exercises 23–38.

28.f(x)=x4+1,x=2

Short Answer

Expert verified

f'(2)=32.

Step by step solution

01

Part (a) Step 1. Given information.

A function is given asf((x)=x4+1and x=c=2.

02

Part (a) Step 2. Solve f'(2) using h→0 definition of the derivative.

We have

f'(c)=limh→0f(c+h)-f(c)hf'(2)=limh→0f(2+h)-f(2)h=limh→0(2+h)4+1-[24+1]h=limh→0(2+h)4+1-24-1h=limh→0[(2+h)2]2-(22)2h=limh→0[(2+h)2-22][(2+h)2+22]h=limh→0(h2+4h)(h2+4h+8)h=limh→0(h+4)(h2+4h+8)=limh→0(h3+4h2+4h2+16h+8h+32)=limh→0(h3+8h2+24h+32)=0+0+0+32=32

03

Part (b) Step 1. Solve f'(2) using z→2 definition of the derivative.

We have

f'(c)=limz→cf(z)-f(c)z-cf'(2)=limz→2f(z)-f(2)z-2=limz→2z4+1-(24+1)z-2=limz→2z4-24z-2=limz→2(z2-22)(z2+22)z-2=limz→2(z-2)(z+2)(z2+4)z-2=limz→2(z+2)(z2+4)=limz→2(z3+4z+2z2+8)=23+4(2)+2(2)2+8=8+8+8+8=32

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