Question

Compute the first thrre derivatives of \\(f(x)=2x^{5}+x^{\\frac{3}{2}}-\\frac{1}{2x}\\)

Answer 1

 Answer

$f'''(x)=120x^2-{\frac{3}{8}}x^{-\frac{3}{2}}+3x^{-4}.$

Explanation

According the linearity property of the derivative for any triple of constants c₁, c₂, c₃ and three functions f, g and h we have

(c₁f + c₂g + c₃h)’ = c₁f’ + c₂g’ + c₃h’.

On the other hand the differentiation rule of power function can be written in the form

$(x^{n})'=nx^{n-1}$

Because,$\frac{1}{x}=x^{-1}$ we can write

$f(x)=2x^5+x^{\frac{3}{2}}-\frac{1}{2}x^{-1}$

So,

$f'(x)=(2x^5+x^{\frac{3}{2}}-\frac{1}{2}x^{-1})'=10x^4+{\frac{3}{2}}x^{\frac{1}{2}}+\frac{1}{2}x^{-2}$

For the second derivative we get

$f''(x)=(10x^4+{\frac{3}{2}}x^{\frac{1}{2}}+\frac{1}{2}x^{-2})'=40x^3+{\frac{3}{4}}x^{-\frac{1}{2}}-x^{-3}$

And for the third derivative we have

$f'''(x)=(40x^3+{\frac{3}{4}}x^{-\frac{1}{2}}-x^{-3})'=120x^2-{\frac{3}{8}}x^{-\frac{3}{2}}+3x^{-4}.$