## Calculus: Early Transcendentals 8th Edition

Consider $x=f(t)=(t-1)^{3}$ and $y=g(t)=(t-1)^{2}$ Thus $g'(t)=2(t-1)$ Now put $t=1$ $g'(1)=2(1-1)=0$ which is not equal to $1$ as per the given statement, that is, $g'(1)\ne 1$. Hence , the statement is false.