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Question 1 of 5
Find the derivative
`y=4-3e^(-x)`
Incorrect
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0:00
Substitute the components into the formula
Differentiating constants makes them `0`
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
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`=` |
$$4-\color{#9a00c7}{f'(-x)}\cdot 3e^{\color{#D800AD}{-x}}$$ |
Substitute known values |
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`=` |
$$0-(\color{#9a00c7}{-1}\cdot 3e^{-x})$$ |
Differentiate `-x` and `4` |
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`y’` |
`=` |
`-3e^(-x)` |
`d/dx (e^(f(x)))=y’` |
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Question 2 of 5
Find the derivative
`y=6e^(x/2)-3e^(-2x)`
Incorrect
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0:00
Substitute the components into the formula
First Term:
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
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`=` |
$$\left(\color{#9a00c7}{f’\left(\frac{x}{2}\right)}\cdot 6e^{\color{#D800AD}{\frac{x}{2}}}\right)-3e^{-2x}$$ |
Substitute known values |
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`=` |
$$\left(\color{#9a00c7}{\frac{1}{2}}\cdot 6e^{\frac{x}{2}}\right)-3e^{-2x}$$ |
Diferrentiate `x/2` |
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`=` |
`3e^(x/2)-3e^(-2x)` |
Second Term:
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`=` |
$$3e^\frac{x}{2}-\left(\color{#9a00c7}{f'(-2x)}\cdot 3e^{\color{#D800AD}{-2x}}\right)$$ |
Substitute known values |
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`=` |
$$3e^\frac{x}{2}-\left(\color{#9a00c7}{-2}\cdot 3e^{-2x}\right)$$ |
Diferrentiate `-2x` |
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`y’` |
`=` |
`3e^(x/2)+6e^(-2x)` |
`d/dx (e^(f(x)))=y’` |
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Question 3 of 5
Find the derivative
`y=(e^(2x)+e^(-x))/2`
Incorrect
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0:00
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 formula
First Term:
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
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`=` |
$$\left(\color{#9a00c7}{f'(2x)}\cdot \frac{e^{\color{#D800AD}{2x}}}{2}\right)+\frac{e^{-x}}{2}$$ |
Substitute known values |
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`=` |
$$\left(\color{#9a00c7}{2}\cdot \frac{e^{2x}}{2}\right)+\frac{e^{-x}}{2}$$ |
Diferrentiate `2x` |
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`=` |
`e^(2x)+(e^(-x))/2` |
`2/2=1` |
Second Term:
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`=` |
$$e^{2x}+\left(\color{#9a00c7}{f'(-x)}\cdot \frac{e^{\color{#D800AD}{-x}}}{2}\right)$$ |
Substitute known values |
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`=` |
$$e^{2x}+\left(\color{#9a00c7}{-1}\cdot \frac{e^{-x}}{2}\right)$$ |
Differentiate `-x` |
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`y’` |
`=` |
`e^(2x)-(e^(-x))/2` |
`d/dx (e^(f(x)))=y’` |
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Question 4 of 5
Find the derivative
`y=e^(3-2x^2)`
Incorrect
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Progress: 0%
0:00
Substitute the components into the formula
Differentiating constants makes them `0`
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
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`=` |
$$\color{#9a00c7}{f'(3-2x^{2})}\cdot e^{\color{#D800AD}{3-2x^{2}}}$$ |
Substitute known values |
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`=` |
$$\color{#9a00c7}{-4x}\cdot e^{3-2x^{2}}$$ |
Diferrentiate `3-2x^2` |
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`y’` |
`=` |
`-4xe^(3-2x^2)` |
`d/dx (e^(f(x)))=y’` |
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Question 5 of 5
Find the derivative
`y=e^(-0.03x)`
Incorrect
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Progress: 0%
0:00
Substitute the components into the formula
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ |
`=` |
$$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ |
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`=` |
$$\color{#9a00c7}{f'(-0.03x)}\cdot e^{\color{#D800AD}{-0.03x}}$$ |
Substitute known values |
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`=` |
$$\color{#9a00c7}{-0.03}\cdot e^{-0.03x}$$ |
Diferrentiate `-0.03x` |
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`y’` |
`=` |
`-0.03e^(-0.03x)` |
`d/dx (e^(f(x)))=y’` |