What is (\partial)/(\partial r)(2t+r^2cos(t)) ?

The answer to (\partial)/(\partial r)(2t+r^2cos(t)) is 2rcos(t)

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(\partial)/(\partial r)(2t+r^2cos(t))

\frac{\partial}{\partial r} (2 t + r^{2} cos (t))

2 r cos (t)

Solution steps

\frac{\partial}{\partial r} (2 t + r^{2} cos (t))

Treat

t

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial r} (2 t) = 0

\frac{\partial}{\partial r} (r^{2} cos (t)) = 2 r cos (t)

= 0 + 2 r cos (t)

Simplify

= 2 r cos (t)

\int \frac{e ^{4 x}}{1 + e ^{2 x}} d x

\int_{0}^{\infty} \frac{x}{e ^{x}} d x

\int_{1}^{e} ln (4 x) d x

(1 - x^{2}) \frac{d y}{d x} = x e^{y}

x \to + 5 lim (\frac{( x + 1 )}{x - 5})

What is (\partial)/(\partial r)(2t+r^2cos(t)) ?
The answer to (\partial)/(\partial r)(2t+r^2cos(t)) is 2rcos(t)