Question
Fill in the blanks:
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)=$ _______.
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)=$ _______.
✓
Answer
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)=\frac{-1}{\sqrt{2}}.$
Solution:
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)$
$\because\ 0<\text{x}<\frac{\pi}{2},\cos\text{x}>0.$
$\text{f(x)}=+\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=(-\sin\text{x})$
$\Rightarrow\ \text{f}'\Big(\frac{\pi}{4}\Big)=-\sin\frac{\pi}{4}=\frac{-1}{\sqrt{2}}$ $\Big[\because\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$
Solution:
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)$
$\because\ 0<\text{x}<\frac{\pi}{2},\cos\text{x}>0.$
$\text{f(x)}=+\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=(-\sin\text{x})$
$\Rightarrow\ \text{f}'\Big(\frac{\pi}{4}\Big)=-\sin\frac{\pi}{4}=\frac{-1}{\sqrt{2}}$ $\Big[\because\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$
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