Answer
$$0.005012,\ \ 0.000012,\ \ 0.25\%$$
Work Step by Step
Given $$f(x)=\tan\frac{x+\pi}{4}, \quad a=0, \quad \Delta x=0.01$$ Since $$ f'(x) = \frac{1}{4}\sec^2\frac{x+\pi}{4} ,\ \ \ f'(0)=0.5 $$ Then \begin{align*} \Delta f &\approx f^{\prime}(a) \Delta x\\ &=(0.5)(0.01)\\ &= 0.005 \end{align*} Since the change $\Delta f$ is given by \begin{align*} \Delta f&=f(a+\Delta x)-f(a)\\ &=f(0.01)-f(0)\\ & \approx0.005012 \end{align*} to find error $$|0.005012-0.005|=0.000012$$ and the percentage $$\frac{0.000012}{0.005012} \times 100 \% \approx0.25\%$$
