Product Rule
- Thread starter Jason76
- Start date
g(x) does not appear in the problem.Given \(\displaystyle (f)(g)\).
\(\displaystyle g'(f) - f'(g)\).
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f'(x) = [4(\dfrac{d}{dx} x)](5x^{2}) + [5(\dfrac{d}{dx} x^{2})](4x)\).
\(\displaystyle f'(x) = [4 (1)](5x^{2}) + [5 (2x)](4x)\).
\(\displaystyle f'(x) = (4)(5x^{2}) + (10x)(4x)\). OK this far - but what happened to the "10"?
\(\displaystyle f'(x) = (4)(5x^{2}) + 4x^{2}\).SIMPLIFY
The question seems to be to show that if you differentiate f(x) as a product,
......\(\displaystyle f(x) = (5x^2)\times (4x)\)
using the product rule,
then you must get the same result you would be differentiating the product
......\(\displaystyle f(x) = 20\ x^3\)
Post Edited
Given \(\displaystyle (f)(g)\).
\(\displaystyle g(f') + f(g')\).
Ex 1.
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f'(x) = (4x)[5(\dfrac{d}{dx} x^{2} )] + (5x^{2})[4(\dfrac{d}{dx} x)]\).
\(\displaystyle f'(x) = (4x)[5(2x)] + (5x^{2})[4(1)]\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + 20x^{2} \).
\(\displaystyle f'(x) = 60x^{2}\).
OR using the power rule
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f(x) = 20x^{3}\).
\(\displaystyle f'(x) = 20 (\dfrac{d}{dx} x^{3})\).
\(\displaystyle f'(x) = 20(3x^{2})\).
\(\displaystyle f'(x) = 60x^{2}\)
Given \(\displaystyle (f)(g)\).
\(\displaystyle g(f') + f(g')\).
Ex 1.
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f'(x) = (4x)[5(\dfrac{d}{dx} x^{2} )] + (5x^{2})[4(\dfrac{d}{dx} x)]\).
\(\displaystyle f'(x) = (4x)[5(2x)] + (5x^{2})[4(1)]\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + 20x^{2} \).
\(\displaystyle f'(x) = 60x^{2}\).
OR using the power rule
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f(x) = 20x^{3}\).
\(\displaystyle f'(x) = 20 (\dfrac{d}{dx} x^{3})\).
\(\displaystyle f'(x) = 20(3x^{2})\).
\(\displaystyle f'(x) = 60x^{2}\)
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\(\displaystyleGiven \(\displaystyle (f)(g)\).
By any chance do you mean: \(\displaystyle Given\ h(x) = f(x) * g(x),\ find\ h'(x).\)
\(\displaystyle g'(f) - f'(g)\). This is WRONG. Not a minus, a plus.
\(\displaystyle f(x) = 5x^{2}(4x)\).
Now you have changed your notation so that what originally was a component function is now the resultant function. So let's clear that up. Please be careful.
\(\displaystyle g(x) = 5x^2 \implies g'(x) = 10x.\)
\(\displaystyle h(x) = 4x \implies h'(x) = 4\)
\(\displaystyle f(x) = g(x) * h(x) \implies f'(x) = g'(x) * h(x) + g(x) * h'(x) = 10x * 4x + 5x^2 * 4 = 40x^2 + 20x^2 = 60x^2.
The whole point of learning things like the product rule and quotient rule is so that you can break a complicated function into simpler functions and work with the pieces.
\)\(\displaystyle
\(\displaystyle f'(x) = (4(\dfrac{d}{dx}[x]))(5x^{2}) + (5(\dfrac{d}{dx}[x^{2}]))(4x)\).
\(\displaystyle f'(x) = (4(1))(5x^{2}) + (5(2x)(4x)\).
\(\displaystyle f'(x) = (4)(5x^{2}) + (10x)(4x)\).
\(\displaystyle f'(x) = (4)(5x^{2}) + 4x^{2}\).
Be careful and pause to think.\)
Given \(\displaystyle (f)(g)\).
\(\displaystyle g(f') + f(g')\).
Ex 1.
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f'(x) = (4x)[5(\dfrac{d}{dx} x^{2} )] + (5x^{2})[4(\dfrac{d}{dx} x)]\).
\(\displaystyle f'(x) = (4x)[5(2x)] + (5x^{2})[4(1)]\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = (4x)(10x) + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + (5x^{2})(4)\).
\(\displaystyle f'(x) = 40x^{2} + 20x^{2} \).
\(\displaystyle f'(x) = 60x^{2}\).
OR using the power rule
\(\displaystyle f(x) = 5x^{2}(4x)\).
\(\displaystyle f(x) = 20x^{3}\).
\(\displaystyle f'(x) = 20 (\dfrac{d}{dx} x^{3})\).
\(\displaystyle f'(x) = 20(3x^{2})\).
\(\displaystyle f'(x) = 60x^{2}\)
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