- #1
The Bob
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Quick question.
To differentiate [tex]y = 20x (2x - 1)^6[/tex] I need to use the product rule:
[tex] y = uv \ \Rightarrow \ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}[/tex]
So that u = 20x and v = (2x-1)6
However, do I differentiate v so that it equals 6(2x-1)5 or do I use:
[tex]y = [f(x)]^n \ \Rightarrow \ \frac{dy}{dx} = n[f(x)]^{n-1} f'(x)[/tex]
to give 12(2x-1)5?
Cheers.
The Bob (2004 ©)
To differentiate [tex]y = 20x (2x - 1)^6[/tex] I need to use the product rule:
[tex] y = uv \ \Rightarrow \ \frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}[/tex]
So that u = 20x and v = (2x-1)6
However, do I differentiate v so that it equals 6(2x-1)5 or do I use:
[tex]y = [f(x)]^n \ \Rightarrow \ \frac{dy}{dx} = n[f(x)]^{n-1} f'(x)[/tex]
to give 12(2x-1)5?
Cheers.
The Bob (2004 ©)
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