Answer :
Answer:
Required differential (a) [tex]dy=-0.0002\times10^{3}[/tex] and (b) [tex]\Delta y=-0.3370[/tex].
Step-by-step explanation:
Given,
[tex]y=\tan x[/tex]
Let, [tex]y=f(x)=\tsn x\hfill (1)[/tex]
(a) To find differential,
[tex]dy[/tex] differentiate (1) we get,
[tex]dy=\sec^2 x dx[/tex]
when, [tex]x=\frac{\pi}{4}[/tex] and dx=-0.1, then
[tex]dy=\sec^2(\frac{\pi}{4})\times (-0.1)=-0.2[/tex]
[tex]dy=-0.0002\times 10^{3}[/tex] (Correct upto four decimal places)
(b) To find,
[tex]\Delta y=\Delta f(x)=f(x+h)-f(x)[/tex] ( where h=dy)
when [tex]x=\frac{\pi}{4}, dx=h=-0.1[/tex]
[tex]\therefore \Delta y=\tan (\frac{\pi}{4}-0.2)-\tan \frac{\pi}{4}[/tex]
[tex]=-0.3370[/tex]
Correct upto four decimal places.
The expression dy in terms of dx is sec²xdx
The value of dy is -0.2
The value of Δy to four decimal places is -0.3374
Given the expression y = tan x
a) On differentiating the given function with respect to x;
[tex]\frac{dy}{dx} = sec^2x\\dy = sec^2x dx[/tex]
b) Substitutung the following parameters
x = π/4 and dx = −0.1
[tex]dy = sec^2(\frac{\pi}{4} )(-0.1)\\dy=\frac{1}{cos^2(\pi/4)} (-0.1)\\dy=\frac{1}{(1/\sqrt{2} )^2}(-0.1)\\dy=2(-0.1)\\dy=-0.2[/tex]
Get Δy;
[tex]\triangle y=f(x+dy)-f(x)\\f(x) = tanx\\f(x+dy) = tan(x+dy)\\\triangle y=tan(x+dy)-tanx\\[/tex]
Given that;
dy = -0.2
x = π/4
[tex]\triangle y=tan(\frac{\pi}{4}+(-0.2) )-tan\frac{\pi}{2}\\\triangle y= tan(\frac{\pi}{4}-0.2)-tan\frac{\pi}{2}\\on \ expansion \\ \triangle y=-0.3374[/tex]
Hence the value of Δy to four decimal places is -0.3374
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