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How do you find the derivative of #h(x)=1/2(x^2+1)(5-2x)#?

Answer 1

Charles Arocha

#h'(x)=x*(5-2x)-(x^2+2)=5x-2x^2-x^2-2=-3x^2+5x-2#

show below

#h(x)=1/2(x^2+1)(5-2x)#

now we will dirvite it

#h'(x)=1/2*[(2x+0)*(5-2x)+(x^2+2)(0-2)]#

#h'(x)=1/2*[2x*(5-2x)+(x^2+2)(-2)]#

#h'(x)=x*(5-2x)-(x^2+2)=5x-2x^2-x^2-2=-3x^2+5x-2#

note that

#y=g(x)*f(x)#

#y'=g(x)*f'(x)+g'(x)*f(x)#

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Answer 2

Paisley Johnson

To find the derivative of ( h(x) = \frac{1}{2}(x^2+1)(5-2x) ), you can use the product rule. The product rule states that if ( f(x) ) and ( g(x) ) are differentiable functions, then the derivative of their product is given by ( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) ). Applying this rule:

Let ( f(x) = \frac{1}{2}(x^2+1) ) and ( g(x) = (5-2x) ).

Now, find the derivatives of ( f(x) ) and ( g(x) ):

( f'(x) = \frac{d}{dx} \left(\frac{1}{2}(x^2+1)\right) = \frac{1}{2} \cdot 2x = x )
( g'(x) = \frac{d}{dx} (5-2x) = -2 )

Now, apply the product rule: [ h'(x) = f'(x)g(x) + f(x)g'(x) ] [ = x(5-2x) + \frac{1}{2}(x^2+1)(-2) ] [ = 5x - 2x^2 - x^2 - 1 ] [ = 5x - 3x^2 - 1 ]

So, the derivative of ( h(x) ) is ( h'(x) = 5x - 3x^2 - 1 ).

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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.