009A Sample Midterm 2, Problem 3

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

Foundations:
Limit Definition of Derivative
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$

Solution:

Step 1:
Let $f(x)={\frac {1+x}{3x}}.$
Using the limit definition of derivative, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Step 2:
Now, we get a common denominator for the fractions in the numerator.
Hence, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}=\frac{-1}{3x^2}}

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