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Local Linear Approximation; Differentials Solutions To Selected ...

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(d) y = x √ 1 − x = x(1 − x) 1/2<br />

(e)<br />

dy = d(x(1 − x) 1/2 )<br />

= x d(1 − x) 1/2 + (1 − x) 1/2 dx<br />

�<br />

1<br />

= x<br />

2 (1 − x)−1/2 �<br />

(−1)dx + (1 − x) 1/2 dx<br />

x<br />

= −<br />

2 √ 1 − x dx + √ 1 − x dx<br />

�<br />

√1 x<br />

= − x −<br />

2 √ �<br />

dx<br />

1 − x<br />

�<br />

√1 2<br />

= − x · √ 1 − x<br />

2 √ 1 − x −<br />

x<br />

2 √ �<br />

dx<br />

1 − x<br />

�<br />

2(1 − x) − x<br />

=<br />

2 √ �<br />

dx<br />

x<br />

�<br />

2 − 3x<br />

=<br />

2 √ �<br />

dx<br />

1 − x<br />

y = (1 + x) −17<br />

dy = d(1 + x) −17<br />

= −17(1 + x) −18 dx<br />

= − 17<br />

dx<br />

(1 + x) 18<br />

7

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