Answer
An approximation for $f′(1)=-1.997$
The actual value of $f′(1)=-2$
The two values are comparable. Therefore, we confirm both the approximation and the actual value of f′(1)
Work Step by Step
We approximate $f'(1)$ using the difference quotient $\frac{f(1+h)-f(1)}{h}$ for values of $h$ near $0$. In this case, we choose $h=0.001$.
Approximate value of $f'(1) = \frac{f(1+h)-f(1)}{h}=\frac{f(1+0.001)-f(1)}{0.001} = \frac{\frac{1}{(1.001)^2}-\frac{1}{1^2}}{0.001} = -1.997$
We apply the power rule to take the derivative and find the real value $f'(1)$:
Real value of $f'(1) = -2x^{-3} = -2(1)^{-3}= -2$
