What is (\partial ^2)/(\partial r\partial θ)(r(θ)sin(θ)) ?

The answer to (\partial ^2)/(\partial r\partial θ)(r(θ)sin(θ)) is sin(θ)+θcos(θ)

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(\partial ^2)/(\partial r\partial θ)(r(θ)sin(θ))

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) sin (θ))

sin (θ) + θ cos (θ)

Solution steps

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) sin (θ))

\frac{\partial}{\partial r} (r θ sin (θ)) = θ sin (θ)

= \frac{\partial}{\partial θ} (θ sin (θ))

\frac{\partial}{\partial θ} (θ sin (θ)) = sin (θ) + θ cos (θ)

= sin (θ) + θ cos (θ)

\int_{- 1}^{1} (x^{2} - 2 x) d x

\frac{d}{d x} (\frac{sin ( 2 x )}{cos ( x ^{2} )})

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

\frac{1}{4 s ^{3}} - \frac{5}{3 s ^{2}}

t

What is (\partial ^2)/(\partial r\partial θ)(r(θ)sin(θ)) ?
The answer to (\partial ^2)/(\partial r\partial θ)(r(θ)sin(θ)) is sin(θ)+θcos(θ)