Question

Find the derivative of the algebraic function. \(f(x)=\frac{c^{2}-x^{2}}{c^{2}+x^{2}}, \quad c\) is a constant

Short answer

The derivative of the function \(f(x) = \frac{c^{2}-x^{2}}{c^{2}+x^{2}}\) is \(f'(x) = \frac{-4c^{2}x}{(c^{2}+x^{2})^{2}}\).

Step-by-step solution

Identify the functions

Firstly, identify two functions \(u(x)\) and \(v(x)\) where:\(u(x)=c^{2}-x^{2}\) and \(v(x)=c^{2}+x^{2}\)

Find the derivatives of the functions

Find the derivatives of these two functions:\(u'(x)=\frac{d}{dx}(c^{2}-x^{2})=-2x\), \(v'(x)=\frac{d}{dx}(c^{2}+x^{2})=2x\)

Apply the quotient rule

Apply the quotient rule which is \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^{2}}\). Put the values and get the derivative: \(f'(x)= \frac{(c^{2}+x^{2})(-2x)-(c^{2}-x^{2})(2x)}{(c^{2}+x^{2})^{2}}\)

Simplify the derivative

Now simplify the derivative obtained, hence:\(f'(x)= \frac{-2c^{2}x-2x^{3} - 2c^{2}x +2x^{3}}{(c^{2}+x^{2})^{2}}\)Further simplifying gives:\(f'(x)= \frac{-4c^{2}x}{(c^{2}+x^{2})^{2}}\)