Differentials Solution 11

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SOLUTION 11: Begin with the function

$f(x)= \arctan x$
and choose

$x-values: 1 \rightarrow 1.1$
so that

$\Delta x = 1.1-1 = 0.1$
The derivative of

$\ y=f(x) \$ is

$f'(x)= \displaystyle{ 1 \over 1+x^2 }$
The exact change of

$y-$ values is

$\Delta y = f(1.1) - f(1)$

$= \arctan(1.1) - \arctan(1)$

$= \arctan(1.1) - \displaystyle{ \pi \over 4 }$

$\approx \arctan(1.1) - 0.7854$
The Differential is

$dy = f'(1) \ \Delta x$

$= \displaystyle{ 1 \over 1+(1)^2 } \cdot (0.1)$

$= \displaystyle{ 1 \over 2 } (0.1)$

$= 0.05$
We will assume that

$\Delta y \approx dy \ \ \ \ \longrightarrow$

$\arctan(1.1) - \displaystyle{ \pi \over 4 } \approx 0.05 \ \ \ \ \longrightarrow$

$\arctan(1.1) \approx \displaystyle{ \pi \over 4 } + 0.05 \ \ \ \ \longrightarrow$

$\arctan(1.1) \approx 0.7854 + 0.05 \ \ \ \ \longrightarrow$

$\arctan(1.1) \approx 0.8354$

NOTE: The number 1 was chosen for its proximity to 1.1 and for it's convenient inverse tangent. Check the accuracy of the final estimate using a CALCULATOR:

$\arctan(1.1) \approx 0.8330$

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