# How do you use the product rule to differentiate #f(x)=e^x(sqrtx+5x^3)#?

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To differentiate ( f(x) = e^x(\sqrt{x} + 5x^3) ) using the product rule, you apply the formula:

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

where ( u = e^x ) and ( v = \sqrt{x} + 5x^3 ).

First, find the derivatives of ( u ) and ( v ):

[ u' = e^x ] [ v' = \frac{1}{2\sqrt{x}} + 15x^2 ]

Now, apply the product rule:

[ f'(x) = e^x(\sqrt{x} + 5x^3)' + e^x(\sqrt{x} + 5x^3)'] [ = e^x \cdot \frac{1}{2\sqrt{x}} + e^x \cdot 15x^2 + e^x \cdot \sqrt{x} + 5e^x \cdot 3x^2 ] [ = \frac{e^x}{2\sqrt{x}} + 15xe^x + e^x\sqrt{x} + 15x^2e^x ]

So, the derivative of ( f(x) ) with respect to ( x ) is:

[ f'(x) = \frac{e^x}{2\sqrt{x}} + 15xe^x + e^x\sqrt{x} + 15x^2e^x ]

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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.

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