How to Simplify Derivatives Using Common Denominators

[bugatti79]

Folks,
Just struggling to see how this is simplified.

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

if we let $$a=(1+f'(x)^2)^{1/2}$$ then I get as far as$$f''(x)[a^{-1/2}-f'(x)^2a^{-3/2}]=f''(x)[a^{-1/2}-f'(x)^2 a^{-1/2} a^{-1}]$$...

 

[Sudharaka]

Folks,
Just struggling to see how this is simplified.

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

if we let $$a=(1+f'(x)^2)^{1/2}$$ then I get as far as$$f''(x)[a^{-1/2}-f'(x)^2a^{-3/2}]=f''(x)[a^{-1/2}-f'(x)^2 a^{-1/2} a^{-1}]$$...

Hi bugatti79, :)

Take the common denominator. The common denominator of $\frac{f''(x)}{((1+f'(x)^2)^{1/2}}-\frac{f'(x)^2 f''(x)}{((1+f'(x)^2)^{3/2}}$ is $(1+f'(x)^2)^{3/2}$ and therefore you have to multiply both the numerator and the denominator of the first fraction by $(1+f'(x)^2)$.