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Calculus1 20 Online
OpenStudy (anonymous):

how does y=x^2(x^3-1) become 5x^4-2x using derivatives?

OpenStudy (anonymous):

Do you know the power rule, or even the product rule for derivatives?

OpenStudy (anonymous):

I vaguely understand it, I get confuse

OpenStudy (anonymous):

Well, you can either use product rule or multiply them out and do power rule. This is the power rule: \[\frac{d}{dx}cx^n=cnx^{n-1}\]To use this, we just: \[x^2(x^3-1)= x^5-x^2 \\ \frac{d}{dx}(x^5-x^2) =\frac{d}{dx}(x^5)-\frac{d}{dx}(x^2)=5x^4-2x \] Then we have product rule: \[\frac{d}{dx}(f\cdot g) =\frac{d}{dx}(f)g\cdot f\frac{d}{dx}(f)\]To use this we say \[f=x^2, g=x^3-1\\ \frac{d}{dx}x^2(x^3-1)=\frac{d}{dx}(x^2)\cdot (x^2-1) + x^2 \frac{d}{dx}(x^3-1) = 5x^4-2x \]

OpenStudy (anonymous):

thank you

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