### Rules of differentiation: Applications of derivatives

### Tangent lines revisited

At the introduction (see *introduction* and *the notion of difference quotient*) of differentiation we started with the question: can we find the slope of a tangent? Now we know how to do it.

Let #f# be a function that is differentiable at #a#. The tangent line to #f# at #a# is given by the linear equation \[y-f(a)=f'(a)\cdot (x-a)\tiny.\]

The point #\rv{x,y}=\rv{a,f(a)}# is a solution of the linear equation. This means that #\rv{a,f(a)}# lies on the line given by the equation.

Moreover, the slope #f'(a)# of #f# at #a# is equal to the slope of that line. Thus, the line given by the linear equation is the unique line through #\rv{a,f(a)}# with slope #f'(a)#. This implies that it is the tangent line to #f# at #a#.

It is not necessary to memorize this formula for the tangent line. What matters is that it is the unique line through the point #\rv{a,f(a)}# with slope #f'(a)#.

This ensures that we can calculate the tangent ourselves.

Determine the

*function rule*#l(x)# for the tangent at the point #\rv{3,{{49}\over{10}}}#. Enter your answer in the form #a\cdot x+b# for suitable values of #a# and #b#.

The requested function rule has the form #l(x)=a\cdot x+b#, where #a# and #b# are real numbers.

The number #a# is the slope of the tangent line, so #a=f'(3)#. The derivative of the function #f# equals #f'(x)={{x}\over{5}}#. Therefore, #a=f'(3)={{3}\over{5}}#.

We conclude that the function rule of the tangent line looks like #l(x)={{3}\over{5}} \cdot x +b#, where #b# is yet to be determined. We do so by using the fact that the tangent line moves through the point #\rv{3,{{49}\over{10}}} #. This means that #{{49}\over{10}}={{3}\over{5}} \cdot 3 +b# and, hence, #b={{49}\over{10}}-{{9}\over{5}} = {{31}\over{10}}#.

In conclusion, the formula for the tangent is #l(x)={{3\cdot x}\over{5}}+{{31}\over{10}}#.

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