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How do you differentiate #f(x)=1/sqrt(x-3x^3+5x^5# using the quotient rule?

Answer 1

Aurora Alvarado

#f'(x)=-(1-9x^2+25x^4)/{2(x-3x^3+5x^5)^(3/2)}.#

Let #f(x)=1/sqrtt=t^(-1/2), where, t=x-3x^3+5x^5#.

So, #f(x)=t^-(1/2), t=x-3x^3+5x^5,# meaning that,

#f" is a function of "t, and, t" is of "x#.

We use The Chain Rule in these situations, which stipulates that,

#f'(x)=((df)/dt)(dt/dx)..............(star)#

Knowing that, #d/dt (t^n)=nx^(n-1)," we have,"(df)/dt=-1/2t^(-1/2-1)# &,

#t=x-3x^3+5x^5rArr dt/dx=1-3(3x^2)+5(5x^4)=1-9x^2+25x^4#

Taken together,

#f'(x)=(-1/2t^(-3/2))(1-9x^2+25x^4), or,#

#f'(x)=-(1-9x^2+25x^4)/{2(x-3x^3+5x^5)^(3/2)}#.

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Answer 2

Ezra Adams

To differentiate ( f(x) = \frac{1}{\sqrt{x-3x^3+5x^5}} ) using the quotient rule, first, identify ( u ) and ( v ) as follows:

( u = 1 )

( v = \sqrt{x-3x^3+5x^5} )

Then, apply the quotient rule formula:

[ f'(x) = \frac{u'v - uv'}{v^2} ]

Where:

( u' = 0 )

( v' = \frac{1}{2\sqrt{x-3x^3+5x^5}} \cdot (1 - 9x^2 + 25x^4) )

After substituting these values into the quotient rule formula, simplify to get the derivative of ( f(x) ).

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Answer from HIX Tutor

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