How do you differentiate #f(x) = (2x+1)(4-x^2)(1+x^2) #?
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The product rule for 3 factors is:
So
Simplify algebraically as you see fit.
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To differentiate the function ( f(x) = (2x+1)(4-x^2)(1+x^2) ), you can use the product rule of differentiation. The product rule states that if ( u(x) ) and ( v(x) ) are differentiable functions of ( x ), then the derivative of their product is given by ( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) ). Applying this rule to ( f(x) ), we get:
[ f'(x) = (2x+1)\frac{d}{dx}[(4-x^2)(1+x^2)] + (4-x^2)(1+x^2)\frac{d}{dx}(2x+1) ]
[ = (2x+1)[(4-x^2)'(1+x^2) + (4-x^2)(1+x^2)'] + (4-x^2)(1+x^2)(2) ]
[ = (2x+1)[(-2x)(1+x^2) + (4-x^2)(2x)] + 2(4-x^2)(1+x^2) ]
[ = (2x+1)[-2x - 2x^3 + 8x - 2x^3] + 2(4-x^2)(1+x^2) ]
[ = (2x+1)(6x - 4x^3) + 2(4-x^2)(1+x^2) ]
[ = 12x^2 - 8x^4 + 6x - 4x^3 + 8 - 2x^2 ]
[ = -8x^4 - 4x^3 + 10x^2 + 6x + 8 ]
So, the derivative of ( f(x) ) is ( f'(x) = -8x^4 - 4x^3 + 10x^2 + 6x + 8 ).
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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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