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Derivatives of Exponential Functions 1Derivatives of Exponential Functions 1
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Question 1 of 5
1. Question
Find the derivative`y=x^4 e^(-x)`Hint
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Product Rule
$$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$`y’=(dy)/dx``u’=``(du)/dx``v’=``(dv)/dx`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x^4` `u’` `=` `4x^3` Derivative of `v`:`v` `=` `e^(-x)` `v’` `=` `-e^(-x)` Substitute the components into the product rule$$\frac{dy}{dx}$$ `=` $$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$ $$y’$$ `=` $$(\color{#9a00c7}{e^{-x}}\cdot\color{#e65021}{4x^3})+(\color{#00880A}{x^4}\cdot\color{#004ec4}{-e^{-x}})$$ Substitute known values `y’` `=` `4x^3 e^(-x)-x^4 e^(-x)` Evaluate `y’=4x^3 e^(-x)-x^4 e^(-x)` -
Question 2 of 5
2. Question
Find the derivative`y=(x^2-x) e^(2x-1)`Hint
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Product Rule
$$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$`y’=(dy)/dx``u’=``(du)/dx``v’=``(dv)/dx`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x^2-x` `u’` `=` `2x-1` Derivative of `v`:`v` `=` `e^(2x-1)` `v’` `=` `2e^(2x-1)` Substitute the components into the product rule$$\frac{dy}{dx}$$ `=` $$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$ $$y’$$ `=` $$(\color{#9a00c7}{e^{2x-1}}\cdot\color{#e65021}{(2x-1)})+(\color{#00880A}{(x^2-x)}\cdot\color{#004ec4}{2e^{2x-1}})$$ Substitute known values `=` `2xe^(2x-1)-e^(2x-1)+2x^2e^(2x-1)-2xe^(2x-1)` Evaluate `=` `-e^(2x-1)+2x^2e^(2x-1)` `2xe^(2x-1)-2xe^(2x-1)` cancels out `y’=-e^(2x-1)+2x^2e^(2x-1)` -
Question 3 of 5
3. Question
Find the derivative`y=(3-x^2)/(e^2)`Hint
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Excellent!
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``(du)/dx``v’=``(dv)/dx`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `3-x^2` `u’` `=` `-2x` Derivative of `v`:`v` `=` `e^x` `v’` `=` `e^x` Substitute the components into the product rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$ $$y’$$ `=` $$\frac{(\color{#9a00c7}{e^x}\cdot\color{#e65021}{-2x})-(\color{#00880A}{(3-x^2)}\cdot\color{#004ec4}{e^x})}{(\color{#9a00c7}{e^x})^2}$$ Substitute known values `=` `(-2xe^x-[(3-x^2)e^x])/(e^(2x))` Evaluate `y’=(-2xe^x-[(3-x^2)e^x])/(e^(2x))` -
Question 4 of 5
4. Question
Find the derivative`y=(e^x-1)/(e^x+1)`Hint
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Fantastic!
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Product Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``(du)/dx``v’=``(dv)/dx`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `e^x-1` `u’` `=` `e^x` Derivative of `v`:`v` `=` `e^x+1` `v’` `=` `e^x` Substitute the components into the product rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$ $$y’$$ `=` $$\frac{(\color{#9a00c7}{(e^x+1)}\cdot\color{#e65021}{e^x})-(\color{#00880A}{(e^x-1)}\cdot\color{#004ec4}{e^x})}{(\color{#9a00c7}{e^x+1})^2}$$ Substitute known values `=` `(e^(2x)+e^x-e^(2x)+e^x)/((e^x+1)^2)` Evaluate `=` `(2e^x)/((e^x+1)^2)` Simplify `y’=(2e^x)/((e^x+1)^2)` -
Question 5 of 5
5. Question
Find the derivative`y=sqrt(e^(2x)+7)`Hint
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Keep Going!
Incorrect
Chain Rule
$$\frac{dy}{dx}=\color{#004ec4}{\frac{dy}{du}}\cdot\color{#e65021}{\frac{du}{dx}}$$First, find the derivative of `u` and the dervative of `y` with respect to `u`.Derivative of `u`:`u` `=` `e^(2x)+7` `(du)/(dx)` `=` `2e^(2x)` Derivative of `y` with respect to `u`:`y` `=` `u^(1/2)` `(dy)/(du)` `=` `1/2 (e^(2x)+7)^(-1/2)` Substitute the components into the chain rule$$\frac{dy}{dx}$$ `=` $$\color{#004ec4}{\frac{dy}{du}}\cdot\color{#e65021}{\frac{du}{dx}}$$ $$y’$$ `=` $$\color{#004ec4}{\frac{1}{2} (e^{2x}+7)^{-\frac{1}{2}}}\cdot\color{#e65021}{2e^{2x}}$$ Substitute known values `=` `e^(2x)(e^(2x)+7)^(-1/2)` `1/2 times 2=1` `=` `(e^(2x))/((e^(2x)+7)^(1/2))` Reciprocate `(e^(2x)+7)^(-1/2)` `=` `(e^(2x))/(sqrt(e^(2x)+7))` Change the exponent to a surd `y’=(e^(2x))/(sqrt(e^(2x)+7))`