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How do you differentiate #g(x) =x^2 sqrt(4+x)# using the product rule?

Answer 1

Isabella Ackerman

#g' (x)=(5x^2+16x)/(2 sqrt(x+4)#

Use the formula #d/dx(u v)= u (dv)/dx + v(du)/dx#

Given #g(x)=x^2 sqrt(4+x)#

#g' (x)= x^2*1/(2 sqrt(4+x)) *1+sqrt(4+x)*2x#

#g' (x)=x^2/(2 sqrt(x+4))+2x* sqrt(x+4)*(2 sqrt(x+4))/(2 sqrt(x+4))#

#g' (x)=(x^2+4x^2+16x)/(2 sqrt(x+4))#

#g' (x)=(5x^2+16x)/(2 sqrt(x+4)#

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Answer 2

Axel Alvarez

To differentiate ( g(x) = x^2 \sqrt{4+x} ) using the product rule, you would first identify the two functions being multiplied: ( f(x) = x^2 ) and ( h(x) = \sqrt{4+x} ). Then, apply the product rule:

[ g'(x) = f'(x)h(x) + f(x)h'(x) ]

Where: [ f'(x) = 2x ] [ h'(x) = \frac{1}{2\sqrt{4+x}} ]

Now substitute these derivatives into the product rule formula:

[ g'(x) = (2x)(\sqrt{4+x}) + (x^2)\left(\frac{1}{2\sqrt{4+x}}\right) ]

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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.