Product Rule
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Question 1 of 4
1. Question
Find the derivative`y=(3x+4)(x-7)`- `y'=` (6x-17)
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` `=` `3x+4` `u’` `=` `3` Power Rule Derivative of `v`:`v` `=` `x-7` `v’` `=` `1` Power Rule 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’` `=` $$(\color{#9a00c7}{x-7}\cdot\color{#e65021}{3})+(\color{#00880A}{3x+4}\cdot\color{#004ec4}{1})$$ Substitute known values `y’` `=` `3x-21+3x+4` `y’` `=` `6x-17` Combine like terms `y’=6x-17` -
Question 2 of 4
2. Question
Find the derivative`y=x(2x-1)^6`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` `u’` `=` `1` Power Rule Derivative of `v`:`v` `=` `(2x-1)^6` `v’` `=` `12(2x-1)^5` Chain Rule 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’` `=` $$(\color{#9a00c7}{(2x-1)^6}\cdot\color{#e65021}{1})+(\color{#00880A}{x}\cdot\color{#004ec4}{12(2x-1)^5})$$ Substitute known values `y’` `=` `(2x-1)^6+12x(2x-1)^5` `y’` `=` `(2x-1)^5[(2x-1)+12x]` Factorize `y’` `=` `(2x-1)^5(14x-1)` Combine like terms `y’=(2x-1)^5(14x-1)` -
Question 3 of 4
3. Question
Find the derivative`y=4x(3x+1)^3`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` `=` `4x` `u’` `=` `4` Power Rule Derivative of `v`:`v` `=` `(3x+1)^3` `v’` `=` `9(3x+1)^2` Chain Rule 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’` `=` $$(\color{#9a00c7}{(3x+1)^3}\cdot\color{#e65021}{4})+(\color{#00880A}{4x}\cdot\color{#004ec4}{9(3x+1)^2})$$ Substitute known values `y’` `=` `4(3x+1)^3+36x(3x+1)^2` `y’=4(3x+1)^3+36x(3x+1)^2` -
Question 4 of 4
4. Question
Find the derivative`y=x^2(3x+2)^3`Hint
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Incorrect
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` `u’` `=` `2x` Power Rule Derivative of `v`:`v` `=` `(3x+2)^3` `v’` `=` `9(3x+2)^2` Chain Rule 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’` `=` $$(\color{#9a00c7}{(3x+2)^3}\cdot\color{#e65021}{2x})+(\color{#00880A}{x^2}\cdot\color{#004ec4}{9(3x+2)^2})$$ Substitute known values `y’` `=` `2x(3x+2)^3+9x^2(3x+2)^2` `y’` `=` `x(3x+2)^2[2(3x+2)+9x]` Factorize `y’` `=` `(x(3x+2)^2)(6x+4+9x)` Distribute `y’` `=` `(x(3x+2)^2)(15x+4)` Combine like terms `y’=(x(3x+2)^2)(15x+4)`