## University Calculus: Early Transcendentals (3rd Edition)

We are given: $f(x)=x^4$ $\delta f=f(1+0.1)-f(1)$ or $\delta f=f(1.1)-f(1)=1.1^4-1=1.4641-(1)=0.4641$ or $\delta f=0.4641$ differentiation of f(x): $f'(x)=4x^3$ $df=f'(1)\times0.1$ $df=(4)\times0.1=0.4$ $df=0.4$ so $|\delta f-df|=|0.4641-0.4|=|0.0641|=0.0641$ The final answer is: a)0.4641, b)0.4, c)0.0641