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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

= \frac{1}{2 \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}}

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

Simplify

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

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

\int_{0}^{\infty} e^{(- 2 x)} d x

4^{(x^{9})}

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

x \to - \infty lim (3^{- x})

(\frac{11}{6}, - \frac{1}{2}), (\frac{1}{3}, \frac{1}{3})

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