### Rules of differentiation: Applications of derivatives

### Approximation

Assume that we are interested in the numerical value of #\sqrt{1.1}#. We know that the tangent line at #x=1# to the graph of #\sqrt{x}# provides a good approximation. The graph of the square root and the tangent line are drawn below.

If we zoom in on the point #\rv{1,1}#, we see that, at the point #x=1#, the graph of #\sqrt{x}# and the tangent line, which is given by the function #\frac{1}{2}x+\frac{1}{2}#, are very close; see the figure below.

In particular, the value of the function of the tangent line at #x=1.1#, the number #\frac{1}{2} \cdot 1.1+{1}{2}=1.05#, is a good approximation of #\sqrt{1.1}#. This is an example of the following tangent approximation formula.

Tangent approximation formula

Let #a# be a point where the function #f# is differentiable. If #h# is chosen close enough to #0#, then #f(a)+f'(a) \cdot h# is a good approximation of #f(a+h)#. In formula:

\[f(a+h)\approx f(a)+f'(a) \cdot h\tiny.\]

According to the definition of derivative: #f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}#.

This means that: #f'(a)\approx\frac{f(a+h)-f(a)}{h}#.

This gives: #f(a+h)\approx f(a)+f'(a) \cdot h#.

There are good estimations for the deviation from #f(a)+f'(a) \cdot h# compared #f(a+h)#, but we will not go into details here.

Using this formula, we can approximate the value of a differentiable function at points for which we would otherwise have needed a calculator.

We use the tangent approximation formula. It states: #f(a+h)\approx f(a)+f'(a) \cdot h#. Here #f(a)=\sqrt{x}#, #a=1.0#, and #h=0.19#.

First, we calculate the derivative of #f# using the

*power rule for differentiation*: #f'(x)={{1}\over{2\cdot \sqrt{x}}}# . Now we can enter the formula: \[\sqrt{1.19}\approx f(1.0)+f'(1.0)\cdot 0.19 = \sqrt{1.0}+\frac{1}{2 \cdot \sqrt{1.0}} \cdot 0.19\approx 1.0950\tiny.\]

For comparison: the precise approximation of #\sqrt{1.19}# to four decimal places is: #1.0909#.

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