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How do you find the derivative of # [e^(1/2x)]/(2x^3)#?

Answer 1

William Abdalla

Use the rule for the derivative of the products of tho functions:

#d(f*g)/dx = g*df/dx + f * dg/dx#

#(e^(1/2x))/(2x^3) =1/2x^(-3)e^(1/2x)#

#d/dx((e^(1/2x))/(2x^3)) = e^(1/2x)*d((1/2x^(-3)))/dx + 1/2x^(-3)*d(e^(1/2x))/dx = #

#= -3/2x^(-4)e^(1/2x)+1/4x^(-3)e^(1/2x) = e^(1/2x)/(4x^4)(x-6)#

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Answer 2

Emilia Aguilar

To find the derivative of the function ( \frac{e^{\frac{1}{2}x}}{2x^3} ), you can use the quotient rule.

The quotient rule states that for functions ( u(x) ) and ( v(x) ), the derivative of ( \frac{u(x)}{v(x)} ) is given by: [ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ]

In this case, ( u(x) = e^{\frac{1}{2}x} ) and ( v(x) = 2x^3 ).

So, ( u'(x) = \frac{1}{2}e^{\frac{1}{2}x} ) and ( v'(x) = 6x^2 ).

Now, applying the quotient rule: [ \frac{d}{dx} \left( \frac{e^{\frac{1}{2}x}}{2x^3} \right) = \frac{\frac{1}{2}e^{\frac{1}{2}x}(2x^3) - e^{\frac{1}{2}x}(6x^2)}{(2x^3)^2} ]

[ = \frac{xe^{\frac{1}{2}x} - 3e^{\frac{1}{2}x}x^2}{4x^6} ]

[ = \frac{e^{\frac{1}{2}x}(x - 3x^2)}{4x^6} ]

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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.