Question

Find the given higher-order derivative.

f''(x) = 9- 3/x

f'''(x)=

Answer 1

Answer:

$f'''(x)=\frac{3}{x^{2}}$

Step-by-step explanation:

We are given with Second-order derivative of function f(x).

$f''(x)=9-\frac{3}{x}$

We need to find Third-order derivative of the function f(x).

$f''(x)=9-\frac{3}{x}=9-3x^{-1}$

We know that,

f'''(x) = (f''(x))'

So,

$f'''(x)=\frac{\mathrm{d}\,f''(x)}{\mathrm{d} x}$

$f'''(x)=\frac{\mathrm{d}\,(9-3x^{-1})}{\mathrm{d} x}$

$f'''(x)=\frac{\mathrm{d}\,9}{\mathrm{d} x}-\frac{\mathrm{d}\,3x^{-1}}{\mathrm{d} x}$

$f'''(x)=0-3\frac{\mathrm{d}\,x^{-1}}{\mathrm{d} x}$

$f'''(x)=-3(-1)x^{-1-1}$

$f'''(x)=3x^{-2}$

$f'''(x)=\frac{3}{x^{2}}$

Therefore, $f'''(x)=\frac{3}{x^{2}}$