How do you differentiate #((5x+1)^2)(2x-1)#?

Answer 1

Use the Product Rule and Chain Rule:

#f'(x)=(5x+1)^{2}\cdot 2+2(5x+1)\cdot 5\cdot (2x-1)#.

This can be simplified in a couple ways. Probably the most useful way is to leave it in a factored form:

#f'(x)=2(5x+1)(5x+1+10x-5)=2(5x+1)(15x-4)#
That way, you can quickly see the critical points of #f(x)# are #x=-\frac{1}{5}# and #x=\frac{4}{15}#.
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Answer 2

To differentiate the expression ( ((5x + 1)^2)(2x - 1) ) with respect to ( x ), you can use the product rule, which states that if you have two functions ( u(x) ) and ( v(x) ), the derivative of their product is given by:

[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) ]

Applying this rule to ( ((5x + 1)^2)(2x - 1) ):

Let ( u(x) = (5x + 1)^2 ) and ( v(x) = (2x - 1) ).

Then, ( u'(x) = 2(5x + 1)(5) ) by the chain rule, and ( v'(x) = 2 ) since the derivative of ( 2x - 1 ) is simply ( 2 ).

Now, applying the product rule:

[ \frac{d}{dx}[((5x + 1)^2)(2x - 1)] = (2(5x + 1)(5))(2x - 1) + ((5x + 1)^2)(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.

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.

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.

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.

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