Question

f(x)=|cosx| find f'(3π/4)​

Answer 1

Answer:

$\boxed{\sf \: \: f'\left(\dfrac{3\pi}{4}\right) = \dfrac{1}{ \sqrt{2} } \: \: }$

Step-by-step explanation:

Given that,

$\sf \: f(x) = |cosx|$

Let first define the function f(x) as

$\bf\: f(x) = |cosx| = \begin{cases} &\sf{ \: \: \: cosx, \: \: 0 < x < \dfrac{\pi}{2} } \\ \\ &\sf{ - cosx, \: \: \dfrac{\pi}{2} < x < \dfrac{3\pi}{2}}\\ \\ &\sf{ \: \: \: cosx, \: \: \dfrac{3\pi}{2} < x < 2\pi} \end{cases}$

So, from this we get

$\sf \: f(x) \: = - \: cosx$

On differentiating both sides w. r. t. x, we get

$\sf \:\dfrac{d}{dx}f(x) \: =\dfrac{d}{dx}( - \: cosx)$

$\sf \: f'(x) = - \dfrac{d}{dx}cosx$

$\sf \: f'(x) = - ( - sinx)$

$\sf \: f'(x) = sinx$

Thus,

$\sf \: f'\left(\dfrac{3\pi}{4}\right) = sin\left(\dfrac{3\pi}{4}\right) =sin\left(\pi - \dfrac{\pi}{4}\right) = sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{ \sqrt{2} }$

Hence,

$\implies\sf \: f'\left(\dfrac{3\pi}{4}\right) = \dfrac{1}{ \sqrt{2} }$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$