009A Sample Midterm 2, Problem 3 Detailed Solution

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

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Background Information:
Recall
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 f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}

Solution:

Step 1:
Let  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 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
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\frac {(1+x+h)3x}{(3x+3h)(3x)}}-{\frac {(1+x)(3x+3h)}{(3x+3h)(3x)}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {\frac {3x+3x^{2}+3xh-(3x+3h+3x^{2}+3hx)}{(3x+3h)(3x)}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3h}{h(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3}{(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\frac {-3}{(3x)(3x)}}\\&=&\displaystyle {-{\frac {1}{3x^{2}}}.}\end{array}}$

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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