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,{{39}\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,{{39}\over{10}}} #. This means that #{{39}\over{10}}={{3}\over{5}} \cdot 3 +b# and, hence, #b={{39}\over{10}}-{{9}\over{5}} = {{21}\over{10}}#.
In conclusion, the formula for the tangent is #l(x)={{3\cdot x}\over{5}}+{{21}\over{10}}#.
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