Topics
>
Precalculus>
Exponential Functions>
Derivatives of Exponential Functions>
Derivatives of Exponential Functions 3Derivatives of Exponential Functions 3
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 5 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- Answered
- Review
-
Question 1 of 5
1. Question
Find the derivative`y=4-3e^(-x)`Hint
Help VideoCorrect
Great Work!
Incorrect
Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Substitute the components into the formulaDifferentiating constants makes them `0`$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$4-\color{#9a00c7}{f'(-x)}\cdot 3e^{\color{#D800AD}{-x}}$$ Substitute known values `=` $$0-(\color{#9a00c7}{-1}\cdot 3e^{-x})$$ Differentiate `-x` and `4` `y’` `=` `-3e^(-x)` `d/dx (e^(f(x)))=y’` `y’=-3e^(-x)` -
Question 2 of 5
2. Question
Find the derivative`y=6e^(x/2)-3e^(-2x)`Hint
Help VideoCorrect
Keep Going!
Incorrect
Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Substitute the components into the formulaFirst Term:$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\left(\color{#9a00c7}{f’\left(\frac{x}{2}\right)}\cdot 6e^{\color{#D800AD}{\frac{x}{2}}}\right)-3e^{-2x}$$ Substitute known values `=` $$\left(\color{#9a00c7}{\frac{1}{2}}\cdot 6e^{\frac{x}{2}}\right)-3e^{-2x}$$ Diferrentiate `x/2` `=` `3e^(x/2)-3e^(-2x)` Second Term:`=` $$3e^\frac{x}{2}-\left(\color{#9a00c7}{f'(-2x)}\cdot 3e^{\color{#D800AD}{-2x}}\right)$$ Substitute known values `=` $$3e^\frac{x}{2}-\left(\color{#9a00c7}{-2}\cdot 3e^{-2x}\right)$$ Diferrentiate `-2x` `y’` `=` `3e^(x/2)+6e^(-2x)` `d/dx (e^(f(x)))=y’` `y’=3e^(x/2)+6e^(-2x)` -
Question 3 of 5
3. Question
Find the derivative`y=(e^(2x)+e^(-x))/2`Hint
Help VideoCorrect
Excellent!
Incorrect
Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Separate the two terms by giving each a denominator of `2``(e^(2x)+e^(-x))/2` `=` `(e^(2x))/2+(e^(-x))/2` Substitute the components into the formulaFirst Term:$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\left(\color{#9a00c7}{f'(2x)}\cdot \frac{e^{\color{#D800AD}{2x}}}{2}\right)+\frac{e^{-x}}{2}$$ Substitute known values `=` $$\left(\color{#9a00c7}{2}\cdot \frac{e^{2x}}{2}\right)+\frac{e^{-x}}{2}$$ Diferrentiate `2x` `=` `e^(2x)+(e^(-x))/2` `2/2=1` Second Term:`=` $$e^{2x}+\left(\color{#9a00c7}{f'(-x)}\cdot \frac{e^{\color{#D800AD}{-x}}}{2}\right)$$ Substitute known values `=` $$e^{2x}+\left(\color{#9a00c7}{-1}\cdot \frac{e^{-x}}{2}\right)$$ Differentiate `-x` `y’` `=` `e^(2x)-(e^(-x))/2` `d/dx (e^(f(x)))=y’` `y’=e^(2x)-(e^(-x))/2` -
Question 4 of 5
4. Question
Find the derivative`y=e^(3-2x^2)`Hint
Help VideoCorrect
Well Done!
Incorrect
Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Substitute the components into the formulaDifferentiating constants makes them `0`$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(3-2x^{2})}\cdot e^{\color{#D800AD}{3-2x^{2}}}$$ Substitute known values `=` $$\color{#9a00c7}{-4x}\cdot e^{3-2x^{2}}$$ Diferrentiate `3-2x^2` `y’` `=` `-4xe^(3-2x^2)` `d/dx (e^(f(x)))=y’` `y’=-4xe^(3-2x^2)` -
Question 5 of 5
5. Question
Find the derivative`y=e^(-0.03x)`Hint
Help VideoCorrect
Keep Going!
Incorrect
Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Substitute the components into the formula$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(-0.03x)}\cdot e^{\color{#D800AD}{-0.03x}}$$ Substitute known values `=` $$\color{#9a00c7}{-0.03}\cdot e^{-0.03x}$$ Diferrentiate `-0.03x` `y’` `=` `-0.03e^(-0.03x)` `d/dx (e^(f(x)))=y’` `y’=-0.03e^(-0.03x)`