How do you find the derivative of #f(x)=x^3(x^3-5x+10)#?

Answer 1

Emily Ammerman

# f'(x)=2x^2(3x^3-10x+15).#

#f(x)=x^3(x^3-5x+10)=x^6-5x^4+10x^3.#

Knowing that, #(x^n)'=n*x^(n-1),# we have,

#f'(x)=(x^6)'-(5x^4)'+(10x^3#)'#

#=6*x^(6-1)-5(x^4)'+10(x^3)'#

#=6x^5-5(4x^3)+10(3x^2)#

# rArr f'(x)=6x^5-20x^3+30x^2=2x^2(3x^3-10x+15).#

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Answer 2

Cameron Alexander

To find the derivative of ( f(x) = x^3(x^3 - 5x + 10) ), you can use the product rule of differentiation. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product is given by ( u'(x)v(x) + u(x)v'(x) ). Applying this rule to ( f(x) ), where ( u(x) = x^3 ) and ( v(x) = x^3 - 5x + 10 ), we get:

[ f'(x) = u'(x)v(x) + u(x)v'(x) ]

[ f'(x) = (3x^2)(x^3 - 5x + 10) + (x^3)(3x^2 - 5) ]

[ f'(x) = 3x^5 - 15x^3 + 30x^2 + 3x^5 - 5x^3 ]

[ f'(x) = 6x^5 - 20x^3 + 30x^2 ]

So, the derivative of ( f(x) ) with respect to ( x ) is ( 6x^5 - 20x^3 + 30x^2 ).

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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.