What is the derivative of f(x)=((x+1)/(x-3))^2 ?

The derivative of f(x)=((x+1)/(x-3))^2 is -(8(x+1))/((x-3)^3)

What is the first derivative of f(x)=((x+1)/(x-3))^2 ?

The first derivative of f(x)=((x+1)/(x-3))^2 is -(8(x+1))/((x-3)^3)

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derivative of

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

- \frac{8 ( x + 1 )}{( x - 3 ) ^{3}}

Solution steps

\frac{d}{d x} ((\frac{x + 1}{x - 3})^{2})

Apply the chain rule:

2 \cdot \frac{x + 1}{x - 3} \frac{d}{d x} (\frac{x + 1}{x - 3})

= 2 \cdot \frac{x + 1}{x - 3} \frac{d}{d x} (\frac{x + 1}{x - 3})

\frac{d}{d x} (\frac{x + 1}{x - 3}) = - \frac{4}{( x - 3 ) ^{2}}

= 2 \cdot \frac{x + 1}{x - 3} (- \frac{4}{( x - 3 ) ^{2}})

Simplify

2 \cdot \frac{x + 1}{x - 3} (- \frac{4}{( x - 3 ) ^{2}}) : - \frac{8 ( x + 1 )}{( x - 3 ) ^{3}}

= - \frac{8 ( x + 1 )}{( x - 3 ) ^{3}}

\frac{d}{d x} (\frac{2}{x ^{2} + 2})

x \to 0 lim (\frac{x ^{5} - 2}{x ^{4}})

\frac{( e ^{x} )}{5 x ^{2}}

\frac{1}{s ( s + 1 ) ( s + 5 )}

- 3 (t - 1)

What is the derivative of f(x)=((x+1)/(x-3))^2 ?
The derivative of f(x)=((x+1)/(x-3))^2 is -(8(x+1))/((x-3)^3)
What is the first derivative of f(x)=((x+1)/(x-3))^2 ?
The first derivative of f(x)=((x+1)/(x-3))^2 is -(8(x+1))/((x-3)^3)