What is (\partial)/(\partial x)(cos(-t+y+e^x)) ?

The answer to (\partial)/(\partial x)(cos(-t+y+e^x)) is -sin(-t+y+e^x)e^x

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(\partial)/(\partial x)(cos(-t+y+e^x))

\frac{\partial}{\partial x} (cos (- t + y + e^{x}))

- sin (- t + y + e^{x}) e^{x}

Solution steps

\frac{\partial}{\partial x} (cos (- t + y + e^{x}))

Treat

t, y

as constants

Apply the chain rule:

- sin (- t + y + e^{x}) \frac{\partial}{\partial x} (- t + y + e^{x})

= - sin (- t + y + e^{x}) \frac{\partial}{\partial x} (- t + y + e^{x})

\frac{\partial}{\partial x} (- t + y + e^{x}) = e^{x}

= - sin (- t + y + e^{x}) e^{x}

\int x^{5} e^{x^{2}} d x

\frac{d}{d x} (ln (cos (ln (x))))

\frac{d y}{d t} = 3 t e^{- y}

y = x arcsin (2 x)

x (\frac{d y}{d x}) = x^{3} + 16 x^{3} y

What is (\partial)/(\partial x)(cos(-t+y+e^x)) ?
The answer to (\partial)/(\partial x)(cos(-t+y+e^x)) is -sin(-t+y+e^x)e^x