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Derivatives of Exponential Functions>
Derivatives of Exponential Functions 2Derivatives of Exponential Functions 2
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Question 1 of 5
1. Question
Find the derivative`y=e^(2x)`Hint
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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$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(2x)}\cdot e^{\color{#D800AD}{2x}}$$ Substitute known values `=` $$\color{#9a00c7}{2}\cdot e^{\color{#D800AD}{2x}}$$ Differentiate `2x` `y’` `=` `2e^(2x)` `d/dx (e^(f(x)))=y’` `y’=2e^(2x)` -
Question 2 of 5
2. Question
Find the derivative`y=e^(4x)+5`Hint
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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`$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(4x)}\cdot e^{\color{#D800AD}{4x}}+5$$ Substitute known values `=` $$\color{#9a00c7}{4}\cdot e^{4x}+0$$ Differentiate `4x` and `5` `y’` `=` `4e^(4x)` `d/dx (e^(f(x)))=y’` `y’=4e^(4x)` -
Question 3 of 5
3. Question
Find the derivative`y=e^(-3x)`Hint
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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$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(-3x)}\cdot e^{\color{#D800AD}{-3x}}$$ Substitute known values `=` $$\color{#9a00c7}{-3}\cdot e^{-3x}$$ Differentiate `-3x` `y’` `=` `-3e^(-3x)` `d/dx (e^(f(x)))=y’` `y’=-3e^(-3x)` -
Question 4 of 5
4. Question
Find the derivative`y=2e^(-x/2)`Hint
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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$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(-\frac{x}{2})}\cdot 2e^{\color{#D800AD}{-\frac{x}{2}}}$$ 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’` `y’=-e^(-x/2)` -
Question 5 of 5
5. Question
Given that `y=e^(px)`, find `p``(d^2y)/(dx^2)-(dy)/(dx)-6y=0`- `p=` (3, -2) or (-2, 3)
Hint
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`y’=``(dy)/dx``y”=``(d^2y)/(dx^2)`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`.`p=3,-2`