Difference between revisions of "009A Sample Midterm 2, Problem 3"
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&&\\ | &&\\ | ||
& = & \displaystyle{\frac{-3}{(3x)(3x)}}\\ | & = & \displaystyle{\frac{-3}{(3x)(3x)}}\\ | ||
| − | & = & \displaystyle{\frac{ | + | & = & \displaystyle{-\frac{1}{3x^2}.} |
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
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| − | | <math>\frac{dy}{dx}=\frac{ | + | | <math>\frac{dy}{dx}=-\frac{1}{3x^2}</math> |
|- | |- | ||
| | | | ||
|} | |} | ||
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 09:14, 13 March 2017
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
| Foundations: |
|---|
| 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 |
| 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}} |