Quotient Rule
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Question 1 of 5
1. Question
Find the derivative`y=x/(x^2+1)`Hint
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``du``v’=``dv`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x` `u’` `=` `1` Power Rule Derivative of `v`:`v` `=` `x^2+1` `v’` `=` `2x` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$ `y’` `=` $$\frac{(\color{#9a00c7}{x^2+1}\cdot\color{#e65021}{1})-(\color{#00880A}{x}\cdot\color{#004ec4}{2x})}{(\color{#9a00c7}{x^2+1})^2}$$ Substitute known values `=` `(x^2+1-2x^2)/((x^2+1)^2)` Evaluate `=` `(1-x^2)/((x^2+1)^2)` Combine like terms `=` `((1-x)(1+x))/((x^2+1)^2)` Factorize `y’=((1-x)(1+x))/((x^2+1)^2)` -
Question 2 of 5
2. Question
Find the derivative`y=(x^2)/(2x+1)`Hint
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``du``v’=``dv`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x^2` `u’` `=` `2x` Power Rule Derivative of `v`:`v` `=` `2x+1` `v’` `=` `2` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$ $$y’$$ `=` $$\frac{(\color{#9a00c7}{2x+1}\cdot\color{#e65021}{2x})-(\color{#00880A}{x^2}\cdot\color{#004ec4}{2})}{(\color{#9a00c7}{2x+1})^2}$$ Substitute known values `=` `(4x^2+2x-2x^2)/((2x+1)^2)` Evaluate `=` `(2x^2+2x)/((2x+1)^2)` Combine like terms `=` `(2x(x+1))/((2x+1)^2)` Factorize `y’=(2x(x+1))/((2x+1)^2)` -
Question 3 of 5
3. Question
Find the derivative`y=(x^2-4)/(x-1)`Hint
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``du``v’=``dv`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x^2-4` `u’` `=` `2x` Power Rule Derivative of `v`:`v` `=` `x-1` `v’` `=` `1` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$ `y’` `=` $$\frac{(\color{#9a00c7}{x-1}\cdot\color{#e65021}{2x})-(\color{#00880A}{x^2-4}\cdot\color{#004ec4}{1})}{(\color{#9a00c7}{x-1})^2}$$ Substitute known values `=` `(2x(x-1)-x^2+4)/((x-1)^2)` Evaluate `=` `(2x^2-2x+x^2+4)/((x-1)^2)` Distribute `=` `(x^2-2x+4)/((x-1)^2)` Combine like terms `y’=(x^2-2x+4)/((x-1)^2)` -
Question 4 of 5
4. Question
Find the derivative`y=(4x^3-5x+3)/(x^3-1)`Hint
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``du``v’=``dv`First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `4x^3-5x+3` `u’` `=` `12x^2-5` Power Rule Derivative of `v`:`v` `=` `x^3-1` `v’` `=` `3x^2` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$ `y’` `=` $$\frac{(\color{#9a00c7}{x^3-1}\cdot\color{#e65021}{12x^2-5})-(\color{#00880A}{4x^3-5x+3}\cdot\color{#004ec4}{3x^2})}{(\color{#9a00c7}{x^3-1})^2}$$ Substitute known values `=` `(12x^5-5x^3-12x^2+5-12x^5+15x^3-9x^2)/((x^3-1)^2)` Evaluate `=` `(10x^3-21x^2+5)/((x^3-1)^2)` Combine like terms `y’=(10x^3-21x^2+5)/((x^3-1)^2)` -
Question 5 of 5
5. Question
Find the derivative`y=(sqrt(2x+1))/(3x-2)`Hint
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Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$`y’=(dy)/dx``u’=``du``v’=``dv`First, convert the surd into an exponent.`(sqrt(2x+1))/(3x-2)` `=` `((2x+1)^(1/2))/(3x-2)` Next, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `(2x+1)^(1/2)` `u’` `=` `(2x+1)^(-1/2)` Chain Rule Derivative of `v`:`v` `=` `3x-2` `v’` `=` `3` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{du}-\color{#00880A}{u}\color{#004ec4}{dv}}{\color{#9a00c7}{v}^2}$$ `y’` `=` $$\frac{(\color{#9a00c7}{3x-2}\cdot\color{#e65021}{(2x+1)^{-\frac{1}{2}}})-(\color{#00880A}{(2x+1)^{\frac{1}{2}}}\cdot\color{#004ec4}{3})}{(\color{#9a00c7}{3x-2})^2}$$ Substitute known values `=` `(((3x-2)(2x+1)^(-1/2))-3((2x+1)^(1/2)))/((3x-2)^2)` Further simplify the numerator`((3x-2)(2x+1)^(-1/2))-3((2x+1)^(1/2))` `=` `(3x-2)/(sqrt(2x+1))-3sqrt(2x+1)` Convert the exponents into surds `=` `((3x-2)-3(2x+1))/(sqrt(2x+1))` Find the common denominator `=` `(3x-2-6x-3)/(sqrt(2x+1))` Distribute `=` `(-3x-5)/(sqrt(2x+1))` Combine like terms Finally, combine the simplified numerator to the denominator.`(-3x-5)/(sqrt(2x+1))xx 1/((3x-2)^2)` `=` `(-3x-5)/(sqrt(2x+1)(3x-2)^2)` `y’=(-3x-5)/(sqrt(2x+1)(3x-2)^2)`
Quizzes
- Power Rule 1
- Power Rule 2
- Power Rule 3
- Power Rule 4
- Chain Rule 1
- Chain Rule 2
- Product Rule
- Quotient Rule
- Derivatives of Exponential Functions 1
- Derivatives of Exponential Functions 2
- Derivatives of Exponential Functions 3
- Derivatives of Trigonometric Functions 1
- Derivatives of Trigonometric Functions 2
- Derivatives of Trigonometric Functions 3