If f(x)=3x2+4x+5, what is f′(1)?
We have f′(x)=2×3x+1×4=6x+4, which implies f′(1)=10. □
If f(x)=−3x4+7x3−6x2−πx+5, what is dxdf(x)?
We have
dxd(−3x4+7x3−6x2−πx+5)=(−3)(4)x4−1+(7)(3)x3−1−(6)(2)x2−1−(π)(1)x1−1+0=−12x3+21x2−12x−π. □
If f(x)=(3x−2)4, what is dxdf(x)?
Note that f(x) is a polynomial but is not in the form given in the summary above. We will later see methods to differentiate this function directly, but to use the tools we have so far, we first expand the polynomial by the binomial theorem. This gives
f(x)=(3x−2)4=(3x)4+4(3x)3(−2)+24⋅3(3x)2(−2)2+1⋅2⋅34⋅3⋅2(3x)(−2)3+(−2)4=81x4−216x3+216x2−96x+16,
which implies
dxdf(x)=(81)(4)x3−(216)(3)x2+(216)(2)x−96=324x3−648x2+432x−96. □
If f(x)=x3, what is h→0limhf(2+h)−f(2)?
Using the above formula, we have
h→0limhf(2+h)−f(2)=h→0limh(2+h)3−23=3×23−1=12. □
If h→0limhf(x+h)−f(x)=4x2, what is f′(2)?
Using the above formula, we know h→0limhf(x+h)−f(x)=dxdf(x)=f′(x)=4x2. Then
f′(2)=h→0limhf(2+h)−f(2)=4×22=16. □
If f(x)=x2+4x, what is h→3limh−3f(h)−f(3)?
Note that h→3limh−3f(h)−f(3) is not in the form given in the formula above. Thus, we use the substitution method in order to get a general formula. Substituting k for h−3, we have
h→3limh−3f(h)−f(3)=k→0limkf(3+k)−f(3).
Then since h→0limhf(x+h)−f(x)=dxdf(x)=f′(x)=2x+4, we have
k→0limkf(3+k)−f(3)=f′(3)=2×3+4=10. □
Find the derivative of f(x)=(ax+b)n for x>−ab.
Write u=ax+b⟹u′=a so that f(x)=un.
Then f′(x)=dxd(un)=nun−1⋅u′=an(ax+b)n−1. □