Question

Let $f,g$ and $h$ be functions from $R$ to $R$. Show that
$(f+g)oh=foh+goh$
$(f\cdot g)oh=\left(foh\right)\cdot \left(goh\right)$

Answer 1

$Letf\left(x\right)=x,g\left(x\right)=\sin x,h\left(x\right)=\log x(f+g)ohf+g=f\left(x\right)+g\left(x\right)=x+\sin x(f+g)=x+\sin x(f+g)h\left(x\right)=h\left(x\right)+\sin h\left(x\right)(f+g)oh=\log x+\sin \log xfoh+gohf\left(x\right)+g\left(x\right)=x+\sin xf[h\left(x\right)+g\left(h\left(x\right)\right)]=h\left(x\right)+\sin h\left(x\right)foh+goh=\log x+\sin \log x\therefore (f+g)oh=foh+goh$

$Letf\left(x\right)=x,g\left(x\right)=\sin x,h\left(x\right)=\log x(f\cdot g)ohf\cdot g=f\left(x\right)\cdot g\left(x\right)=x\sin x(f\cdot g)=x\sin x(f\cdot g)h\left(x\right)=h\left(x\right)\cdot \sin h\left(x\right)(f\cdot g)oh=\log x\cdot \sin \log x(f\cdot g)oh=\log x\cdot \sin \log xf\left(x\right)\cdot g\left(x\right)=x\cdot \sin xf[h\left(x\right)\cdot g\left(h\left(x\right)\right)]=h\left(x\right)\cdot \sin h\left(x\right)foh\cdot goh=\log x\cdot \sin \log x\therefore (f\cdot g)oh=foh\cdot goh$