"Higher-order derivatives Find f′(x),f″(x), and f‴(x).f(x)=x2(2+x−3)" was answered by our experts

floymdiT

Answered question

2021-10-16

Higher-order derivatives Find $f^{'} (x), f^{″} (x),$ and $f^{‴} (x)$ .
$f (x) = x^{2} (2 + x^{- 3})$

insonsipthinye

Skilled2021-10-17Added 83 answers

Step 1
Given,

f (x) = x^{2} (2 + x^{- 3})

Step 2
Consider,

f (x) = x^{2} (2 + x^{- 3})

f (x) = 2 x^{2} + x^{2} x^{- 3}

f (x) = 2 x^{2} + x^{2 - 3}

f (x) = 2 x^{2} + x^{- 1}

Now differentiate with respect to "x" we get,
Use the power rule of differentiation we get,

\frac{d}{d x} x^{n} = n x^{n - 1}

f^{'} (x) = 2 (2 x^{2 - 1}) + (- 1 x^{- 1 - 1})

f^{'} (x) = 4 x - x^{- 2}

Now again differentiate with respect to "x" we get,

f^{″} (x) = 4 - (- 2 x^{- 2 - 1})

f^{″} (x) = 4 + 2 x^{- 3}

Now again differentiate with respect to "x" we get,

f^{‴} (x) = 0 + 2 (- 3 x^{- 3 - 1})

f^{‴} (x) = - 6 x^{- 4}

Therefore,

f^{'} (x) = 4 x - x^{- 2}

f^{″} (x) = 4 + 2 x^{- 3}

f^{‴} (x) = - 6 x^{- 4}

Jeffrey Jordon

Expert2022-08-15Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?