Difference between revisions of "009A Sample Midterm 2, Problem 4"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
Line 9: | Line 9: | ||
!Foundations: | !Foundations: | ||
|- | |- | ||
− | | | + | |'''1.''' Product Rule |
− | |||
− | |||
− | |||
|- | |- | ||
− | | | + | |'''2.''' Quotient Rule |
− | |||
|} | |} | ||
Line 25: | Line 21: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
− | | | + | |Using the Product Rule, we have |
|- | |- | ||
− | | | + | | <math>f'(x)=x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).</math> |
|} | |} | ||
Line 33: | Line 29: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |Now, we have |
|- | |- | ||
− | | | + | | <math>\begin{array}{rcl} |
+ | \displaystyle{f'(x)} & = & \displaystyle{x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1)}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1).} | ||
+ | \end{array}</math> | ||
|} | |} | ||
Line 67: | Line 67: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | |'''(a)''' | + | | '''(a)''' <math>x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math> |
|- | |- | ||
|'''(b)''' | |'''(b)''' | ||
|} | |} | ||
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 13:19, 17 February 2017
Find the derivatives of the following functions. Do not simplify.
- a)
- b) where
Foundations: |
---|
1. Product Rule |
2. Quotient Rule |
Solution:
(a)
Step 1: |
---|
Using the Product Rule, we have |
Step 2: |
---|
Now, we have |
(b)
Step 1: |
---|
Step 2: |
---|
Final Answer: |
---|
(a) |
(b) |