Information
You have already completed the quiz before. Hence you can not start it again.
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
-
Question 1 of 5
Find the derivative
`y=e^(2x)`
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitute the components into the formula
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
|
|
`=` |
$$\color{#9a00c7}{f'(2x)}\cdot e^{\color{#D800AD}{2x}}$$ |
Substitute known values |
|
|
`=` |
$$\color{#9a00c7}{2}\cdot e^{\color{#D800AD}{2x}}$$ |
Differentiate `2x` |
|
`y’` |
`=` |
`2e^(2x)` |
`d/dx (e^(f(x)))=y’` |
-
Question 2 of 5
Find the derivative
`y=e^(4x)+5`
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitute the components into the formula
Differentiating constants makes them `0`
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
|
|
`=` |
$$\color{#9a00c7}{f'(4x)}\cdot e^{\color{#D800AD}{4x}}+5$$ |
Substitute known values |
|
`=` |
$$\color{#9a00c7}{4}\cdot e^{4x}+0$$ |
Differentiate `4x` and `5` |
|
`y’` |
`=` |
`4e^(4x)` |
`d/dx (e^(f(x)))=y’` |
-
Question 3 of 5
Find the derivative
`y=e^(-3x)`
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitute the components into the formula
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
|
|
`=` |
$$\color{#9a00c7}{f'(-3x)}\cdot e^{\color{#D800AD}{-3x}}$$ |
Substitute known values |
|
`=` |
$$\color{#9a00c7}{-3}\cdot e^{-3x}$$ |
Differentiate `-3x` |
|
`y’` |
`=` |
`-3e^(-3x)` |
`d/dx (e^(f(x)))=y’` |
-
Question 4 of 5
Find the derivative
`y=2e^(-x/2)`
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitute the components into the formula
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
|
|
`=` |
$$\color{#9a00c7}{f'(-\frac{x}{2})}\cdot 2e^{\color{#D800AD}{-\frac{x}{2}}}$$ |
Substitute known values |
|
|
`=` |
$$\color{#9a00c7}{-\frac{1}{2}}\cdot 2e^{-\frac{x}{2}}$$ |
Differentiate `-x/2` |
|
`y’` |
`=` |
`-e^(-x/2)` |
`d/dx (e^(f(x)))=y’` |
-
Question 5 of 5
Given that `y=e^(px)`, find `p`
`(d^2y)/(dx^2)-(dy)/(dx)-6y=0`
Incorrect
Loaded: 0%
Progress: 0%
0:00
First, find the first and second derivative of `y`
First Derivative:
`y` |
`=` |
`e^(px)` |
`y’` |
`=` |
`pe^(px)` |
Second Derivative:
`y` |
`=` |
`e^(px)` |
`y”` |
`=` |
`p^2e^(px)` |
Substitute the components into the equation
$$\color{#004ec4}{\frac{d^2y}{dx^2}}-\color{#e65021}{\frac{dy}{dx}}-6\color{#00880A}{y}$$ |
`=` |
`0` |
|
$$\color{#004ec4}{p^2e^{px}}-\color{#e65021}{pe^{px}}-6\color{#00880A}{e^{px}}$$ |
`=` |
`0` |
Substitute known values |
`(p^2e^(px)-pe^(px)-6e^(px))``divide e^(px)` |
`=` |
`0``divide e^(px)` |
Divide both sides by `e^(px)` |
`p^2-p-6` |
`=` |
`0` |
Evaluate |
`(p-3)(p+2)` |
`=` |
`0` |
Factorize |
Therefore, the value of `p` is `3` and `-2`.