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How do you differentiate #g(y) =(x^2 + 6)sqrtx # using the product rule?

Answer 1

Elijah Abram

#g'y=2x^(3/2)+(x^2+6)/(2sqrt(x))#

#g(y)=(x^2+6)sqrt(x)=(x^2+6)(x^(1/2))#

Rule of the product:

#f'(uv)=vu'+uv'#

#u=x^2+6, v=x^(1/2)#

#u'=2x, v'=1/2x^(-1/2)=1/(2sqrt(x))#

#g'y=2x^(3/2)+(x^2+6)/(2sqrt(x))#

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Answer 2

Isabella Ackerman

To differentiate ( g(y) = (x^2 + 6)\sqrt{x} ) using the product rule, you first identify the two functions being multiplied together: ( u(x) = x^2 + 6 ) and ( v(x) = \sqrt{x} ). Then, you find the derivatives of these functions, ( u'(x) ) and ( v'(x) ). Afterward, apply the product rule formula:

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

So, the derivative ( g'(y) ) is:

[ g'(y) = (2x)(\sqrt{x}) + (x^2 + 6)\left(\frac{1}{2\sqrt{x}}\right) ]

Simplify this expression to get the final answer.

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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.