009A Sample Midterm 2, Problem 3

Use the definition of the derivative to find   ${\frac {dy}{dx}}$   for the function  $y={\frac {1+x}{3x}}.$

ExpandFoundations:
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

ExpandStep 1:
Let  $f(x)={\frac {1+x}{3x}}.$
Using the limit definition of derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\frac {1+(x+h)}{3(x+h)}})-({\frac {1+x}{3x}})}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\frac {1+x+h}{3x+3h}})-({\frac {1+x}{h}})}{h}}.}\end{array}}$

ExpandStep 2:
Now, we get a common denominator for the fractions in the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\frac {(1+x+h)3x}{(3x+3h)(3x)}}-{\frac {(1+x)(3x+3h)}{(3x+3h)(3x)}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {\frac {3x+3x^{2}+3xh-(3x+3h+3x^{2}+3hx)}{(3x+3h)(3x)}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3h}{h(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3}{(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\frac {-3}{(3x)(3x)}}\\&=&\displaystyle {{\frac {-1}{3x^{2}}}.}\end{array}}$

ExpandFinal Answer:
${\frac {dy}{dx}}={\frac {-1}{3x^{2}}}$

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