What is (\partial)/(\partial s)(-3sqrt(s^2+t^2)) ?

The answer to (\partial)/(\partial s)(-3sqrt(s^2+t^2)) is -(3s)/(sqrt(s^2+t^2))

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(\partial)/(\partial s)(-3sqrt(s^2+t^2))

\frac{\partial}{\partial s} (- 3 s^{2} + t^{2})

- \frac{3 s}{s ^{2} + t ^{2}}

Solution steps

\frac{\partial}{\partial s} (- 3 s^{2} + t^{2})

Treat

t

as a constant

Take the constant out:

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

= - 3 \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

= - 3 \cdot \frac{1}{2 s ^{2} + t ^{2}} \cdot 2 s

Simplify

- 3 \cdot \frac{1}{2 s ^{2} + t ^{2}} \cdot 2 s : - \frac{3 s}{s ^{2} + t ^{2}}

= - \frac{3 s}{s ^{2} + t ^{2}}

y^{^{'}} = y + 3

\int \frac{1}{x ^{2} + 6 x} d x

\frac{d}{d x} (e^{1 x})

f (x) = 7 x - 2 x

\int \frac{2 x ^{2}}{2 x + 1} d x

What is (\partial)/(\partial s)(-3sqrt(s^2+t^2)) ?
The answer to (\partial)/(\partial s)(-3sqrt(s^2+t^2)) is -(3s)/(sqrt(s^2+t^2))