Question

If
$f(x) = \sqrt{1 - sin2x} \: find \: \frac{dy}{dx} \: at \: x = \frac{\pi}{3}$
Differentiate and solve at indicated point​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = \sqrt{1 - sin2x}$

can be rewritten as

$\rm :\longmapsto\:y = \sqrt{ {sin}^{2}x + {cos}^{2}x -2sinxcosx}$

can be further rewritten as

$\rm :\longmapsto\:y = \sqrt{ {(sinx - cosx)}^{2} }$

We know,

$\red{\rm :\longmapsto\: \: \: \: \boxed{\tt{ \: \: \: \: \sqrt{ {x}^{2} } = |x| \: \: \: \: }}}$

So, using this identity, we have

$\rm :\longmapsto\:y = |sinx - cosx|$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ &\sf{ \: \: 0 \: \: when \: x = 0}\\ &\sf{ \: \: x \: \: when \: x > 0} \end{cases}\end{gathered}\end{gathered}$

So, using this, we get

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |sinx - cosx| = \begin{cases} &\sf{ -(sinx - cosx) \: \: when \: x < \dfrac{\pi}{4} } \\ \\ &\sf{ \: \: 0 \: \: when \: x = \dfrac{\pi}{4}}\\ \\ &\sf{ \: \: sinx - cosx \: \: when \: x > \dfrac{\pi}{4}} \end{cases}\end{gathered}\end{gathered}$

So,

$\rm\implies \:When \: x \: = \: \dfrac{\pi}{3}$

then

$\rm :\longmapsto\:y = sinx - cosx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}(sinx - cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = cosx - ( - sinx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = cosx + sinx$

Thus,

$\rm :\longmapsto\:\dfrac{dy}{dx}_{at \: x \: = \: \dfrac{\pi}{3}} = cos\dfrac{\pi}{3} + sin\dfrac{\pi}{3}$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{at \: x \: = \: \dfrac{\pi}{3}} = \dfrac{ \sqrt{3} }{2} + \dfrac{1}{2}$

$\rm\implies \: \: \boxed{\tt{ \dfrac{dy}{dx}_{ \: \: at \: x \: = \: \dfrac{\pi}{3}} \: = \: \dfrac{ \sqrt{3} + 1}{2} \: \: }}$

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More to Know :-

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

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = \sqrt{1 - sin2x}$

can be rewritten as

$\rm :\longmapsto\:y = \sqrt{ {sin}^{2}x + {cos}^{2}x -2sinxcosx}$

can be further rewritten as

$\rm :\longmapsto\:y = \sqrt{ {(sinx - cosx)}^{2} }$

We know,

$\red{\rm :\longmapsto\: \: \: \: \boxed{\tt{ \: \: \: \: \sqrt{ {x}^{2} } = |x| \: \: \: \: }}}$

So, using this identity, we have

$\rm :\longmapsto\:y = |sinx - cosx|$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ &\sf{ \: \: 0 \: \: when \: x = 0}\\ &\sf{ \: \: x \: \: when \: x > 0} \end{cases}\end{gathered}\end{gathered}$

So, using this, we get

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |sinx - cosx| = \begin{cases} &\sf{ -(sinx - cosx) \: \: when \: x < \dfrac{\pi}{4} } \\ \\ &\sf{ \: \: 0 \: \: when \: x = \dfrac{\pi}{4}}\\ \\ &\sf{ \: \: sinx - cosx \: \: when \: x > \dfrac{\pi}{4}} \end{cases}\end{gathered}\end{gathered}$

So,

$\rm\implies \:When \: x \: = \: \dfrac{\pi}{3}$

then

$\rm :\longmapsto\:y = sinx - cosx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}(sinx - cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = cosx - ( - sinx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = cosx + sinx$

Thus,

$\rm :\longmapsto\:\dfrac{dy}{dx}_{at \: x \: = \: \dfrac{\pi}{3}} = cos\dfrac{\pi}{3} + sin\dfrac{\pi}{3}$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{at \: x \: = \: \dfrac{\pi}{3}} = \dfrac{ \sqrt{3} }{2} + \dfrac{1}{2}$

$\rm\implies \: \: \boxed{\tt{ \dfrac{dy}{dx}_{ \: \: at \: x \: = \: \dfrac{\pi}{3}} \: = \: \dfrac{ \sqrt{3} + 1}{2} \: \: }}$

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More to Know :-

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