What is the derivative of e^{x+y}-x ?

The derivative of e^{x+y}-x is e^{x+y}(1+(dy)/(dx))-1

What is the first derivative of e^{x+y}-x ?

The first derivative of e^{x+y}-x is e^{x+y}(1+(dy)/(dx))-1

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\frac{d}{d x} (e^{x + y} - x)

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

Solution steps

\frac{d}{d x} (e^{x + y} - x)

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

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

\frac{d}{d x} (cos (sin (6 x)))

\frac{\partial}{\partial y} (\frac{1 + t ^{2}}{3 y - y ^{2}})

\frac{4 x ^{2} + 6 x + 6}{x}

x \to \frac{π}{2} lim (\frac{1}{cos ( x )})

\frac{d}{d x} (\frac{cos ( 3 ) x ^{6}}{2 x})

What is the derivative of e^{x+y}-x ?
The derivative of e^{x+y}-x is e^{x+y}(1+(dy)/(dx))-1
What is the first derivative of e^{x+y}-x ?
The first derivative of e^{x+y}-x is e^{x+y}(1+(dy)/(dx))-1