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Chapter 4 | Applications of Derivatives
L ( x ) = f (0)+ f ′(0)( x −0).
Because
f ( x ) = (1+ x ) n ⇒ f (0) =1 f ′( x ) = n (1+ x ) n −1 ⇒ f ′(0) = n ,
the linear approximation is given by Figure 4.10 (a).
L ( x ) =1+ n ( x −0) =1+ nx We can approximate (1.01) 3 by evaluating L (0.01) when n =3. We conclude that (1.01) 3 = f (1.01) ≈ L (1.01) = 1 + 3(0.01) = 1.03.
Figure 4.10 (a) The linear approximation of f ( x ) at x =0 is L ( x ). (b) The actual value of 1.01 3 is 1.030301. The linear approximation of f ( x ) at x =0 estimates 1.01 3 to be 1.03.
4.7 Find the linear approximation of f ( x ) = (1+ x ) 4 at x =0 without using the result from the preceding example.
Differentials We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials . Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values. When we first looked at derivatives, we used the Leibniz notation dy / dx to represent the derivative of y with respect to x . Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning to the expressions dy and dx . Suppose y = f ( x ) is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable dy by (4.2) dy = f ′( x ) dx . It is important to notice that dy is a function of both x and dx . The expressions dy and dx are called differentials .We can
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