What is (\partial)/(\partial s)(ln(1+s^2+t^2)) ?

The answer to (\partial)/(\partial s)(ln(1+s^2+t^2)) is (2s)/(1+s^2+t^2)

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(\partial)/(\partial s)(ln(1+s^2+t^2))

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

\frac{2 s}{1 + s ^{2} + t ^{2}}

Solution steps

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

Treat

t

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{1}{z ^{3} z ^{2} - 9} d z

x \to 3 lim (- \frac{5}{8} \cdot x^{4})

y = x^{2} - 6 x, y = - x^{2} + 4 x

y^{^{'}} = - 50 k y + k

cos (θ)

What is (\partial)/(\partial s)(ln(1+s^2+t^2)) ?
The answer to (\partial)/(\partial s)(ln(1+s^2+t^2)) is (2s)/(1+s^2+t^2)