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Derivatives of Exponential Functions>
Derivatives of Exponential Functions 3Derivatives of Exponential Functions 3
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Question 1 of 5
1. Question
Find the derivative`y=4-3e^(-x)`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`$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$4-\color{#9a00c7}{f'(-x)}\cdot 3e^{\color{#D800AD}{-x}}$$ 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
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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 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
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Product Rule with Base “e”
$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$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
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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`$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(3-2x^{2})}\cdot e^{\color{#D800AD}{3-2x^{2}}}$$ 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
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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$$\frac{d}{dx}(e^{\color{#D800AD}{f(x)}})$$ `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `=` $$\color{#9a00c7}{f'(-0.03x)}\cdot e^{\color{#D800AD}{-0.03x}}$$ 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)`