How do you differentiate #f(x)=1/3(2x^3-4)# using the product rule?

Answer 1

Emma Aldridge

#f'(x)=2x^2#

Since there isn't a product of two functions in this case, the product rule is not necessary.

The #color(blue)"product rule"# is used in the following situation.

#f(x)=g(x).h(x)#

here #f(x)=2/3x^3-4/3" distribute the bracket"#

now differentiate using the #color(blue)"power rule"#

That is #color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(ax^n)=nax^(n-1)" and " d/dx("constant")=0)color(white)(a/a)|)))#

#rArrf'(x)=(3xx2/3)x^2-0=2x^2#

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Answer 2

Emily Ammerman

To differentiate ( f(x) = \frac{1}{3}(2x^3 - 4) ) using the product rule, let ( u = \frac{1}{3} ) and ( v = 2x^3 - 4 ). Then, apply the product rule formula: ( (uv)' = u'v + uv' ). Differentiate ( u ) and ( v ) separately to get ( u' ) and ( v' ), then substitute them into the product rule formula.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.