i) Let g(x)=x5. We know that g′(x)=5x4 by Exercise 4 of this chapter. Hence, we can write
f(x)=g(x+3)=(x+3)5and
f′(x)=g′(x+3)=5(x+3)4by Exercise 8a). This also means that
f′(x+3)=5(x+6)4.ii) Let z=x+3. Then x=z−3 and
f(x+3)=f(z)=(z−3)5.By the same argument as in i), we have
f′(z)=5(z−3)4and
f′(z+3)=5z4.iii) Let z=x+3. Then x=z−3 and
f(x+3)=f(z)=(z+2)7.It follows that
f′(z)=7(z+2)6and
f′(z+3)=7(z+5)6.