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Questions and Answers
Given $f(x) = u(x)v(x)w(x)$, which of the following represents $f'(x)$?
Given $f(x) = u(x)v(x)w(x)$, which of the following represents $f'(x)$?
- $u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x)$ (correct)
- $u'(x) + v'(x) + w'(x)$
- $u(x)v(x)w(x) + u'(x)v'(x)w'(x)$
- $u'(x)v'(x)w'(x)$
If $f(2) = -3$, $f'(2) = 4$, $g(2) = 5$, and $g'(2) = -1$, what is $(fg)'(2)$?
If $f(2) = -3$, $f'(2) = 4$, $g(2) = 5$, and $g'(2) = -1$, what is $(fg)'(2)$?
- 19
- 17 (correct)
- -17
- -19
What is the derivative of $h(x) = (3x^2 + 2),\text{cos}(x)$?
What is the derivative of $h(x) = (3x^2 + 2),\text{cos}(x)$?
- $6x \,\text{sin}(x) + (3x^2 + 2)\,\text{cos}(x)$
- $-6x \,\text{sin}(x) - (3x^2 + 2)\,\text{cos}(x)$
- $-6x \,\text{sin}(x) + (3x^2 + 2)\,\text{cos}(x)$
- $6x \,\text{cos}(x) - (3x^2 + 2)\,\text{sin}(x)$ (correct)
Given $y = x^2 ,\text{e}^x$, find $\frac{dy}{dx}$.
Given $y = x^2 ,\text{e}^x$, find $\frac{dy}{dx}$.
Determine the derivative of $f(x) = (x^3 - 1)(2x + 5)$ evaluated at $x = 0$.
Determine the derivative of $f(x) = (x^3 - 1)(2x + 5)$ evaluated at $x = 0$.
Flashcards
Product Rule Formula
Product Rule Formula
The derivative of two functions multiplied together. (f*g)' = f'g + fg'
Product Rule (Three Factors)
Product Rule (Three Factors)
When you have 3 functions multiplied, you take the derivative of the first, times the other two, plus the derivative of the second, times the other two, plus the derivative of the third, times the other two. (fgh)' = f'gh + fg'h + fgh'
How to Use the Product Rule
How to Use the Product Rule
- Find the derivatives of f(x) and g(x). 2. Plug the derivatives and the original functions into: f'(x)g(x) + f(x)g'(x). 3. Simplify.
(fg)'(3)
(fg)'(3)
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Study Notes
Product Rule for Derivatives
- To find the derivative of two functions multiplied together, use the formula: (f*g)' = f'g + fg'
Example 1: x² sin(x)
- f(x) = x²
- g(x) = sin(x)
- f '(x) = 2x
- g'(x) = cos(x)
- (x² sin(x))' = 2x sin(x) + x² cos(x)
- Can be simplified to x (2 sin(x) + x cos(x))
Example 2: (5x - 9x³) (8 + x²)
- f(x) = 5x - 9x³
- g(x) = 8 + x²
- f '(x) = 5 - 27x²
- g'(x) = 2x
- ((5x - 9x³) (8 + x²))' = (5 - 27x²) (8 + x²) + (5x - 9x³) (2x)
Example 3: 4 sin(x) tan(x)
- f(x) = 4 sin(x)
- g(x) = tan(x)
- f '(x) = 4 cos(x)
- g'(x) = sec²(x)
- (4 sin(x) tan(x))' = 4 cos(x) tan(x) + 4 sin(x) sec²(x)
- Simplified: 4 sin(x) sec²(x) [1 + cos(x)]
Example 4: 5x sin(x) - x³ tan(x)
- Product rule is used on each term separately
- First term: 5x sin(x)
- f(x) = 5x
- g(x) = sin(x)
- f '(x) = 5
- g'(x) = cos(x)
- Second term: x³ tan(x)
- h(x) = x³
- k(x) = tan(x)
- h'(x) = 3x²
- k'(x) = sec²(x)
- (5x sin(x) - x³ tan(x))' = 5 sin(x) + 5x cos(x) - [3x² tan(x) + x³ sec²(x)]
Product Rule with Three Factors
- (fgh)' = f'gh + fg'h + fgh'
Example 5: (x² + 6) (7 - 8x) (3 + 5x³)
- f(x) = (x² + 6)
- g(x) = (7 - 8x)
- h(x) = (3 + 5x³)
- f '(x) = 2x
- g'(x) = -8
- h'(x) = 15x²
- ((x² + 6) (7 - 8x) (3 + 5x³))' = 2x(7 - 8x)(3 + 5x³) + (x² + 6)(-8)(3 + 5x³) + (x² + 6)(7 - 8x)(15x²)
Evaluating Derivatives with Given Values
- If f(3) = 4, f'(3) = -7, g(3) = -8, g'(3) = 5, then what is (fg)'(3)?
- (fg)' = f'g + fg'
- (fg)'(3) = f'(3)g(3) + f(3)g'(3)
- (fg)'(3) = (-7)(-8) + (4)(5) = 56 + 20 = 76
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