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Mathematical Expression Editor
We compute the derivative of a quotient.
The Quotient Rule
If and are differentiable then at all points where .
In words, the derivative of a quotient is the bottom times the derivative of the
top minus the top times the derivative of the bottom, all over the bottom
squared.
example 1 Find if We have with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x-1)(1)- (x+1)(1)}{(x-1)^2}\\ &= \frac{(x-1)-(x+1)}{(x-1)^2} \\ &= -\frac{2}{(x-1)^2}. \end{align*}
Here is a video of Example 1
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(problem 1a) Compute
The derivative of is
The derivative of is
Collect like terms in the numerator
The derivative of with respect to is
(problem 1b) Compute
The derivative of is
The derivative of is
Collect like terms in the numerator
The derivative of with respect to is
(problem 1c) Compute
The derivative of is
The derivative of is
Collect like terms in the numerator
The derivative of with respect to is
example 2 Find if We write with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x^3 + 5)(2x-3) - (x^2-3x)(3x^2)}{(x^3 + 5)^2}\\ &= \frac{(2x^4 -3x^3 + 10x - 15) -(3x^4-9x^3)}{(x^3 + 5)^2}\\ &= \frac{2x^4 -3x^3 + 10x - 15 -3x^4+9x^3}{(x^3 + 5)^2}\\ &= \frac{-x^4 +6x^3 + 10x - 15}{(x^3 + 5)^2} \end{align*}
Here is a video of Example 2
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(problem 2a) Compute
The derivative of is
The derivative of is
Collect like terms in the numerator
The derivative of with respect to is
(problem 2b) Compute
The derivative of is
The derivative of is
Collect like terms in the numerator
The numerator simplifies to
The derivative of with respect to is
example 3 Find if We write with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(2x+1)(-\sin (x))- \cos (x)\cdot 2}{(2x+1)^2}\\ &= -\frac{(2x+1)\sin (x) + 2\cos (x)}{(2x+1)^2} \end{align*}
Here is a video of Example 3
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(problem 3) Compute
The derivative of with respect to is
example 4 Find if We write with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x^2+1)e^x - e^x(2x)}{(x^2 + 1)^2} \\ &= \frac{(x^2 - 2x +1)e^x}{(x^2 + 1)^2} \\ &= \frac{e^x(x-1)^2}{(x^2 + 1)^2} \end{align*}
Here is a video of Example 4
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(problem 4)
example 5 Find if We write with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{(x+1)(1/x) - \ln (x)}{(x+1)^2}\\ &= \frac{(x+1)(1/x) - \ln (x)}{(x+1)^2}\cdot \frac{x}{x}\\ &= \frac{(x+1) - x\ln (x)}{x(x+1)^2}. \end{align*}
Here is a video of Example 5
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(problem 5)
example 6 Find if First rewrite as Then with: To use the quotient rule we need the derivatives: We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\cos (x)\cos (x) - \sin (x)(-\sin (x))}{\cos ^2(x)}\\ &= \frac{\cos ^2(x)+ \sin ^2(x)}{\cos ^2(x)}\\ &= \frac{1}{\cos ^2(x)} = \sec ^2(x) \end{align*}
This important formula is worth remembering: if then .
Here is a video of Example 6
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(problem 6a)
(problem 6b)
(problem 6c) Find the equation of the tangent line to at . The equation is
example 7 Find if . First rewrite as and then with: To use the quotient rule we need the derivatives:
We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\sin (x)(-\sin (x)) -\cos (x)\cos (x) }{\sin ^2(x)}\\ &= -\frac{\sin ^2(x)+ \cos ^2(x)}{\sin ^2(x)}\\ &= -\frac{1}{\sin ^2(x)} = -\csc ^2(x) \end{align*}
This important formula is worth remembering: if then .
Here is a video of Example 7
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(problem 7a)
(problem 7b)
example 8 Find if First we rewrite as and then with: To use the quotient rule we need the
derivatives: We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\cos (x)\cdot 0 - 1\cdot (-\sin (x))}{\cos ^2(x)}\\ &= \frac{ \sin (x)}{\cos ^2(x)} \\ &= \frac{ 1}{\cos (x)} \cdot \frac{\sin (x)}{\cos (x)}\\ &= \sec (x) \tan (x) \end{align*}
This important formula is worth remembering: if then .
Here is a video of Example 8
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(problem 8a)
(problem 8b)
(problem 8c) Find the equation of the tangent line to at . The equation is
example 9 Find if First we rewrite as and then with: To use the quotient rule we need the
derivatives: We can now write: \begin{align*} h'(x) &= \frac{g(x)f'(x) - f(x)g'(x)}{g^2(x)}\\ &= \frac{\sin (x)\cdot 0 - 1\cdot (\cos (x))}{\sin ^2(x)}\\ &= \frac{ -\cos (x)}{\sin ^2(x)} \\ &= -\frac{ 1}{\sin (x)} \cdot \frac{\cos (x)}{\sin (x)}\\ &= -\csc (x) \cot (x) \end{align*}
This important formula is worth remembering: if then .
(problem 9a)
(problem 9b)
(problem 9c) Find the equation of the tangent line to at . The equation is
example 10 Find if We could use the quotient rule, but it is easier to rewrite as
and find using the power and constant rules. We have:
(problem 10a) Find if Rewrite as
(problem 10b) Find if Rewrite as
Here is a detailed, lecture style video on the Quotient Rule: