If y-fx-sin x- then differential dy- where x- 6 -dx-0-01

The correct option is A √( 3 )( 0.005 ) f(x)=sin x ∴ f'(x)=cos x ∴ f'( π #160; 6 #160; ) =cosπ #160; 6 #160; =√( 3 ) #160; 2 N ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Applications of Differential Equations

If y=fx=sin...

Question

If $y = f (x) = sin x$ , then differential $d y = . . . . . . . . . . . .$ , where $x = \frac{π}{6}, d x = 0.01$

\sqrt{3} (0.005)

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\sqrt{3} (0.05)

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\sqrt{3} (0.5)

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\sqrt{3}

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Solution

The correct option is A $\sqrt{3} (0.005)$
$f (x) = sin x$
$∴ f^{'} (x) = cos x ∴ f^{'} (\frac{π}{6}) = cos \frac{π}{6} = \frac{\sqrt{3}}{2}$
Now $d y = f^{'} (x) . d x = \frac{\sqrt{3}}{2} (0.01) = \frac{\sqrt{3}}{2} (0.005)$

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If y=f(x)=sinx, then differential dy=............, where x=π6,dx=0.01

The correct option is A √3(0.005)f(x)=sinx∴f′(x)=cosx∴f′(π6)=cosπ6=√32Now dy=f′(x).dx=√32(0.01)=√32(0.005)

If $y = f (x) = sin x$ , then differential $d y = . . . . . . . . . . . .$ , where $x = \frac{π}{6}, d x = 0.01$

The correct option is A $\sqrt{3} (0.005)$
$f (x) = sin x$
$∴ f^{'} (x) = cos x ∴ f^{'} (\frac{π}{6}) = cos \frac{π}{6} = \frac{\sqrt{3}}{2}$
Now $d y = f^{'} (x) . d x = \frac{\sqrt{3}}{2} (0.01) = \frac{\sqrt{3}}{2} (0.005)$