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

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Answer 1 · 2017-01-30T10:23:59

#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#.

In such cases, we apply The Chain Rule, which states 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#

Altogether,

#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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