anonymous
  • anonymous
How do I differentiate this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ 1 }{ f \sqrt[]{x} }\]
SolomonZelman
  • SolomonZelman
Wait, maybe it is: \(\large\color{black}{ \displaystyle \frac{ 1}{\sqrt{x} } }\) ?
anonymous
  • anonymous
The question says what is the derivative of \[g(x)=\frac{ 1 }{ f(\sqrt{x}) }\]

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anonymous
  • anonymous
suppose \(f(x)=\sin(x)\) making \(f(\sqrt{x})=\sin(\sqrt{x})\) how would you take the derivative of \[\frac{1}{\sin(\sqrt{x})}\]?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle g(x)=\left\{~f(h(x))~\right\}^n }\) (when n is not 0) First, derivative of \(x^n\) is \(nx^{n-1}\). \(\large\color{black}{ \displaystyle n\left\{~f(h(x))~\right\}^{n-1} }\) then you apply the chain rule, knowing that the derivative of f(x) is f'(x) you get: \(\large\color{black}{ \displaystyle n\left\{~f(h(x))~\right\}^{n-1} \times f'(h(x)) }\) and then once again you apply the chain rule knowing that the derivative of h(x) is h'(x). \(\large\color{black}{ \displaystyle g'(x)= n\left\{~f(h(x))~\right\}^{n-1} \times f'(h(x)) \times h'(x) }\)
phi
  • phi
you might want to first write the expression as \[ g(x) = \left( f\left(x^\frac{1}{2}\right)\right)^{-1} \] now use Solomon's ideas

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