What is the derivative of f(s)=sqrt((s^2+1)/(s^2+4)) ?

The derivative of f(s)=sqrt((s^2+1)/(s^2+4)) is (3s)/(sqrt(s^2+1)(s^2+4)^{3/2)}

What is the first derivative of f(s)=sqrt((s^2+1)/(s^2+4)) ?

The first derivative of f(s)=sqrt((s^2+1)/(s^2+4)) is (3s)/(sqrt(s^2+1)(s^2+4)^{3/2)}

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derivative of

f (s) = \frac{s ^{2} + 1}{s ^{2} + 4}

\frac{3 s}{s ^{2} + 1 ( s ^{2} + 4 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

Apply the chain rule:

\frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

= \frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4})

\frac{d}{d s} (\frac{s ^{2} + 1}{s ^{2} + 4}) = \frac{6 s}{( s ^{2} + 4 ) ^{2}}

= \frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \cdot \frac{6 s}{( s ^{2} + 4 ) ^{2}}

Simplify

\frac{1}{2 \frac{s ^{2} + 1}{s ^{2} + 4}} \cdot \frac{6 s}{( s ^{2} + 4 ) ^{2}} : \frac{3 s}{s ^{2} + 1 ( s ^{2} + 4 ) ^{\frac{3}{2}}}

= \frac{3 s}{s ^{2} + 1 ( s ^{2} + 4 ) ^{\frac{3}{2}}}

\int cos^{4} (x) d x

\frac{\partial}{\partial y} (x y^{2})

x \to 0 + lim (e^{- x})

\frac{\partial}{\partial x} ((x^{2} + 5 x y - 2 y)^{8})

e^{- 5 t} sin (7 t)

What is the derivative of f(s)=sqrt((s^2+1)/(s^2+4)) ?
The derivative of f(s)=sqrt((s^2+1)/(s^2+4)) is (3s)/(sqrt(s^2+1)(s^2+4)^{3/2)}
What is the first derivative of f(s)=sqrt((s^2+1)/(s^2+4)) ?
The first derivative of f(s)=sqrt((s^2+1)/(s^2+4)) is (3s)/(sqrt(s^2+1)(s^2+4)^{3/2)}