Difference between revisions of "009A Sample Midterm 1, Problem 4"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 64: | Line 64: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |Using the Quotient Rule, we have |
|- | |- | ||
| − | | | + | | <math>h'(x)=\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}.</math> |
| − | |||
| − | |||
| − | |||
| − | |||
|} | |} | ||
| Line 76: | Line 72: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, using the Chain Rule, we have |
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| + | \displaystyle{h'(x)} & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-5x^3)'-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(x^2+1)'}{(\sqrt{x^2+1})^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 92: | Line 90: | ||
| '''(b)''' <math>\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math> | | '''(b)''' <math>\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math> | ||
|- | |- | ||
| − | |'''(c)''' | + | | '''(c)''' <math>\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}</math> |
|} | |} | ||
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 10:46, 16 February 2017
Find the derivatives of the following functions. Do not simplify.
- a)
- b) where
- c) 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 h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}}
| Foundations: |
|---|
| 1. Product Rule |
| 2. Quotient Rule |
| 3. Chain Rule |
Solution:
(a)
| Step 1: |
|---|
| Using the Product Rule, 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 f'(x)=(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'.} |
| Step 2: |
|---|
| Now, 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{(\sqrt{x})'(x^2+2)+\sqrt{x}(x^2+2)'}\\ &&\\ & = & \displaystyle{\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x).} \end{array}} |
(b)
| Step 1: |
|---|
| Using the Quotient Rule, 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 g'(x)=\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}.} |
| Step 2: |
|---|
| Now, 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{g'(x)} & = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}.} \end{array}} |
(c)
| Step 1: |
|---|
| Using the Quotient Rule, 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 h'(x)=\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}.} |
| Step 2: |
|---|
| Now, using the Chain Rule, 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{h'(x)} & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}}\\ &&\\ & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-5x^3)'-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(x^2+1)'}{(\sqrt{x^2+1})^2}}\\ &&\\ & = & \displaystyle{\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}.} \end{array}} |
| Final Answer: |
|---|
| (a) 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 \bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x)} |
| (b) 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{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}} |
| (c) 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{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}} |