Statement Compute the derivatives with the chain rule: (a) p(x)=(x2+1)5p(x)=(x^2+1)^5p(x)=(x2+1)5, then p′(1)p'(1)p′(1) (b) q(x)=4x−3q(x)=\sqrt{4x-3}q(x)=4x−3 (c) r(x)=(1+ex)3r(x)=(1+e^{x})^3r(x)=(1+ex)3 Solution (a) p′(x)=5(x2+1)4⋅2x=10x(x2+1)4p'(x)=5(x^2+1)^4\cdot 2x=10x(x^2+1)^4p′(x)=5(x2+1)4⋅2x=10x(x2+1)4. Hence p′(1)=10⋅1⋅24=160p'(1)=10\cdot 1\cdot 2^4=160p′(1)=10⋅1⋅24=160. (b) q(x)=(4x−3)1/2⇒q′(x)=12(4x−3)−1/2⋅4=24x−3q(x)=(4x-3)^{1/2}\Rightarrow q'(x)=\dfrac12(4x-3)^{-1/2}\cdot 4=\dfrac{2}{\sqrt{4x-3}}q(x)=(4x−3)1/2⇒q′(x)=21(4x−3)−1/2⋅4=4x−32. (c) r′(x)=3(1+ex)2⋅ex=3ex(1+ex)2r'(x)=3(1+e^{x})^2\cdot e^{x}=3e^{x}(1+e^{x})^2r′(x)=3(1+ex)2⋅ex=3ex(1+ex)2. p′=10x(x2+1)4, p′(1)=160; q′=24x−3; r′=3ex(1+ex)2\boxed{p'=10x(x^2+1)^4,\ p'(1)=160;\ q'=\tfrac{2}{\sqrt{4x-3}};\ r'=3e^x(1+e^x)^2}p′=10x(x2+1)4, p′(1)=160; q′=4x−32; r′=3ex(1+ex)2