What is (\partial)/(\partial s)(-2sqrt(s^2+t^2)) ?

The answer to (\partial)/(\partial s)(-2sqrt(s^2+t^2)) is -(2s)/(sqrt(s^2+t^2))

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(\partial)/(\partial s)(-2sqrt(s^2+t^2))

\frac{\partial}{\partial s} (- 2 s^{2} + t^{2})

- \frac{2 s}{s ^{2} + t ^{2}}

Solution steps

\frac{\partial}{\partial s} (- 2 s^{2} + t^{2})

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 2 \frac{\partial}{\partial s} (s^{2} + t^{2})

Apply the chain rule:

\frac{1}{2 s ^{2} + t ^{2}} \frac{\partial}{\partial s} (s^{2} + t^{2})

= \frac{1}{2 s ^{2} + t ^{2}} \frac{\partial}{\partial s} (s^{2} + t^{2})

\frac{\partial}{\partial s} (s^{2} + t^{2}) = 2 s

= - 2 \cdot \frac{1}{2 s ^{2} + t ^{2}} \cdot 2 s

Simplify

- 2 \cdot \frac{1}{2 s ^{2} + t ^{2}} \cdot 2 s : - \frac{2 s}{s ^{2} + t ^{2}}

= - \frac{2 s}{s ^{2} + t ^{2}}

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

\frac{d}{d x} (5 arcsin (x^{4}))

\int 23 e^{39 x} d x

\frac{8}{8 x + 9}

\frac{d y}{d x} + 6 y = 8

What is (\partial)/(\partial s)(-2sqrt(s^2+t^2)) ?
The answer to (\partial)/(\partial s)(-2sqrt(s^2+t^2)) is -(2s)/(sqrt(s^2+t^2))